Solution: (d) OA = 40 P
tan 45o
OB = 40 ; AB = OA + OB = 40[1 + 3] 40 m
tan 30o O
= 40[ 3 + 1] = 40 × 2.732 = 109.28 metre. 45o 30o
A B
Example: 22 A tower subtends an angle α at a point A in the plane of its base and the angle of depression of the foot of the tower at a
Solution: (b)
point l meters just above A is β. The height of the tower is [MP PET 1990; RPET 1990]
Example: 23
Solution: (a) (a) l tan β cot α (b) l tan α cot β (c) l tan α tan β (d) l cot α cot β
Example: 24 H = tan α ⇒ H = OA tan α ……..(i) T
Solution: (b) OA
Example: 25 l = tan β ⇒ OA = l cot β ..…..(ii) P H
OA β
l
From (i) and (ii) H = l tan α cot β . α O
A
A tower subtends an angle of 30o at a point distant d from the foot of the tower and on the same level as the foot of the
tower. At a second point h vertically above the first, the depression of the foot of the tower is 60o. The height of the tower
is [MP PET 1993]
(a) h (b) h (c) 3h (d) 3h
3 3d d
Let CD is tower
From ∆BCD, H = tan 30o ........(i) A D
d 60°
and from ∆ABD, h = tan 60o ........(ii) hH
d
30° 60°
H Bd
tan 30o C
= tan 60o h
Divide equation (ii) from equation (i), we have d ⇒ H = 3 .
h
d
A flag-staff of 5m high stands on a building of 25 m high. At an observer at a height of 30m. The flag-staff and the building
subtend equal angles. The distance of the observer from the top of the flag-staff is [EAMCET 1993]
(a) 53 (b) 5 3 (c) 5 2 (d) None of these
2 2 3
We have, tanα = 5 and tan 2α = 30 30m 30m
x x α 3m
α
∴ tan 2α = 5 30 ⇒ tan 2α = 6 tanα 30m
cot α
⇒ 3 − 3 tan2 α = 1 ⇒ tan α = 2 30 m
3
∴ x = 5 cot α = 5 3 . 30m
2
The length of the shadows of a vertical pole of height h, thrown by the sun’s ray at three different moments are h, 2h and
3h. The sum of the angles of elevation of the rays at these three moments is equal to [MP PET 2000]
(a) π (b) π (c) π (d) π
2 3 4 6
P
Solution: (a) tanα = h = 1⇒ α = 45o
Example: 26 h
Solution: (b)
tan β = h ⇒ β = tan−1 1 ; tan γ = h ⇒ γ = tan−1 1 h
Example: 27 2h 2 3h 3
Solution: (c)
Example: 28 ∴ α+β +γ = 45o + tan−1 1 + tan−1 1 = 45o + 45o = π . C γ βα
Solution: (d) 2 3 2
B A 2h h O
3h
A tower subtends angles α, 2α, 3α respectively at points A, B and C, all lying on a horizontal line through the foot of the
tower. Then AB = [EAMCET 2003]
BC
(a) sin 3α (b) 1 + 2 cos 2α (c) 2 + cos 3α (d) sin 2α
sin 2α sinα
From sine rule
⇒ BE BC
sin(180o − 3α) = sin α
⇒ AB = BC (Since BE = AB)
sin 3α sin α
⇒ AB = sin 3α = 3 − 4 sin2 α ⇒ AB 3 − 2(1 − cos 2α) ⇒ AB = 1+ 2cos 2α . 2
BC sinα BC = BC C
The angle of elevation of the top of a tower from a point A due south of the tower is α and from a point B due east of the
tower is β. If AB = d , then the height of the tower is [Roorkee 1979; Haryana CEE 1998]
(a) d (b) d P
tan2 α − tan2 β tan2 α + tan2 β
(c) d (d) d h βB
cot2 α + cot2 β cot2 α − tan2 β d
O
OB = h cot β α
OA = h cot α A
⇒ h2 = cot 2 d2 ⇒ h= d.
β + cot2 α cot2 β + cot2 α
The angular elevation of a tower CD at a point A due south of it is 60o and at a point B due west of A, the elevation is
30o . If AB = 3 km, the height of the tower is [MP PET 1998]
(a) 2 3 km (b) 2 6 km D
(c) 33 km (d) 36 km h
2 4
B 30o
In ∆CBD , tan 30o = h ⇒ BC = 3h , C
BC
3 60o
In ∆ACD , tan 60o = h ⇒ AC = h A
AC 3
Now, AB2 = AC2 + CB2 , 32 = 3h2 + h2 , 27 = 10h2
3
h2 = 27 ⇒h= 33 = 36 ~− 36 km.
10 10 20 4
Example: 29 A pole stands vertically inside a triangular park ∆ABC. If the angle of elevation of the top of the pole from each corner of
Solution: (b)
the park is same, then in ∆ ABC the foot of the pole is at the [IIT Screening 2000]
(a) Centroid (b) Circumcentre Q
(c) Incentre (d) Orthocentre
Let PQ be the pole, since the angle of Q from each of the points A, B, C is the same = θ.
∴ PA = PB = PC = h cot θ h
Since P is equidistant from A, B, C. Aθ θ C
∴ P is circumcentre of ∆ ABC.
Pθ
B
Complex numbers
Introduction.
Number system consists of real numbers (−5,7, 1 , 3..............etc.) and imaginary numbers ( − 5, − 9 ....etc.)
3
If we combine these two numbers by some mathematical operations, the resulting number is known as Complex
Number i.e., “Complex Number is the combination of real and imaginary numbers”.
(1) Basic concepts of complex number
(i) General definition : A number of the form x + iy, where x, y∈R and i = − 1 is called a complex
number so the quantity − 1 is denoted by 'i' called iota thus i = − 1 .
A complex number is usually denoted by z and the set of complex number is denoted by c
i.e., c = {x + i y: x∈R, y∈R,i = − 1}
For example, 5 + 3i, − 1 + i, 0 + 4i, 4 + 0i etc. are complex numbers.
Note : Euler was the first mathematician to introduce the symbol i (iota) for the square root of – 1
with property i 2 = −1. He also called this symbol as the imaginary unit.
• Iota (i) is neither 0, nor greater than 0, nor less than 0.
• The square root of a negative real number is called an imaginary unit.
• For any positive real number a, we have − a = − 1× a = − 1 a = i a
• i − a = − a.
• The property a b = ab is valid only if at least one of a and b is non-negative. If a and b are both
negative then a b = − ab .
• If a < 0 then a = |a|i .
(2) Integral powers of iota (i) : Since i = − 1 hence we have i2 = −1 , i3 = −i and i 4 = 1 . To find the
value of in (n > 4), first divide n by 4. Let q be the quotient and r be the remainder.
i.e., n = 4q + r where 0 ≤ r ≤ 3
in = i 4q+r = (i 4 )q .(i)r = (1)q .(i)r = ir
In general we have the following results i4n = 1, i4n+1 = i, i4n+2 = −1, i4n+3 = −i , where n is any integer.
In other words, in = (−1)n / 2 if n is even integer and in = (−1)n−1 / 2i if n is odd integer.
The value of the negative integral powers of i are found as given below :
i −1 1 i3 = i3 = −i,i−2 1 = 1 = −1, i−3 1 i = i = i, i−4 1 = 1 = 1
= i = i4 = i2 −1 = i3 = i4 1 = i4 1
Important Tips
• The sum of four consecutive powers of i is always zero i.e., in + in+1 + in+2 + in+3 = 0, n ∈ I.
• in = 1, i, − 1, − i, where n is any integer.
• (1 + i)2 = 2i, (1 − i)2 = −2i
• 1+i = i, 1−i = −i, 2i =1−i
1−i 1+i i −1
Example: 1 200 [MP PET 1996]
Solution: (c)
∑If i2 = −1, then the value of in is
n=1
(a) 50 (b) – 50 (c) 0 (d) 100
∑200 = i (1 − i200) (since G.P.) = i (1 − 1) = 0.
in = i + i2 + i3 + ..... + i200 1−i 1−i
n=1
Example: 2 If i = − 1 and n is a positive integer, than in + in+1 + in+2 + in+3 = [Rajasthan PET 2001; Karnataka CET 1994]
Solution: (d)
(a) 1 (b) i (c) in (d) 0
in + in+1 + in+2 + in+3 = in(1 + i + i2 + i3) = in(1 + i − 1 − i) = o.
Trick: Since the sum of four consecutive powers of i is always zero.
⇒ in + in+1 + in+2 + in+3 = 0, n ∈ I.
1 + i x
1 − i
Example: 3 If =1 then [AIEEE 2003; Rajasthan PET 2003]
Solution: (a)
(a) x = 4n, where n is any positive integer (b) x = 2n, where n is any positive integer
(c) x = 4n +1, where n is any positive integer (d) x = 2n +1, where n is any positive integer
1+ i 1 + i 1 + i 1 + i x ix = 1 ⇒
1− i 1 − i 1 + i 1 − i
= . ⇒ =1 ⇒ x = 4n, n ∈ I +.
Example: 4 1 + i2 + i4 + i6 + ..... + i2n is [EAMCET 1980]
Solution: (d) (a) Positive (b) Negative (c) Zero (d) Can not be determined
Example: 5
Solution: (d) n (which is depend upon the value of n).
∑1 + i2 + i4 + i6 + ..... + i2n = i2k = 1 or – 1
k=0
If x = 3 + i, then x3 − 3x2 − 8x + 15 = [UPSEAT 2003]
(a) 6 (b) 10 (c) – 18 (d) – 15
Given that; x − 3 = i ⇒ (x − 3)2 = i2 ⇒ x2 − 6x + 10 = 0
Now, x3 − 3x2 − 8x + 15 = x(x2 − 6x + 10) + 3(x2 − 6x + 10) − 15 = 0 + 0 − 15 = −15.
Example: 6 The complex number 2n + (1 + i)2n , (n ∈ Z) is equal to
(1 − i)2n 2n
(a) 0 (b) 2 (c) [1 + (−1)n].in (d) None of these
Solution: (d) (1 + i)2n = ((1 + i)2)n = (1 + i2 + 2i)n = (1 − 1 + 2i)n = 2nin
(1 − i)2n = ((1 − i)2)n = (1 + i2 − 2i)n = (1 − 1 − 2i)n = (−2)nin
∴ 2n + (1 + i)2n 2n + 2n in 1 + in = 1 + (−1)ni2n = 1 + (−1)n(i2)n
(1 − i)2n 2n = (−2)nin 2n = (−1)nin (−1)n in (−1)n in
= 1 + (−1)n(−1)n = 1 + (−1)2n = 1+1 = 2 .
(−1)n in (−1)n in (−1)n in (−1)n in
Real and Imaginary Parts of a Complex Number.
If x and y are two real numbers, then a number of the form z = x + iy is called a complex number. Here ‘x’ is
called the real part of z and ‘y’ is known as the imaginary part of z. The real part of z is denoted by Re(z) and the
imaginary part by Im(z).
If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4.
Note : A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely
imaginary if its real part is zero i.e., Re(z) = 0.
• i can be denoted by the ordered pair (0,1).
• The complex number (a, b) can also be split as (a, 0) + (0, 1) (b, 0).
Important Tips
• A complex number is an imaginary number if and only if its imaginary part is non-zero. Here real part may or may not be zero.
• All purely imaginary numbers except zero are imaginary numbers but an imaginary number may or not be purely imaginary.
• A real number can be written as a + i.0, therefore every real number can be considered as a complex number whose imaginary part
is zero. Thus the set of real number (R) is a proper subset of the complex number (C) i.e., R ⊂ C.
• Complex number as an ordered pair : A complex number may also be defined as an ordered pair of real numbers and may be
denoted by the symbol (a,b). For a complex number to be uniquely specified, we need two real numbers in particular order.
Algebraic Operations with Complex Numbers.
Let two complex numbers z1 = a + ib and z2 = c + id
Addition : (a + ib) + (c + id) = (a + c) + i(b + d)
Subtraction : (a + ib) − (c + id) = (a − c) + i(b − d)
Multiplication : (a + ib)(c + id) = (ac − bd) + i(ad + bc)
Division : a + ib (when at least one of c and d is non-zero)
c + id
a + ib = (a + ib) . (c − id) (Rationalization)
c + id (c + id) (c − id)
a + ib = (ac + bd) + i(bc − ad) .
c + id c2 + d2 c2 + d2
Properties of algebraic operations with complex numbers : Let z1, z2 and z3 are any complex
numbers then their algebraic operation satisfy following operations:
(i) Addition of complex numbers satisfies the commutative and associative properties
i.e., z1 + z2 = z2 + z1 and (z1 + z2 ) + z3 = z1 + (z2 + z3 ).
(ii) Multiplication of complex number satisfies the commutative and associative properties.
i.e., z1z2 = z2 z1 and (z1z2 )z3 = z1(z2 z3 ).
(iii) Multiplication of complex numbers is distributive over addition
i.e., z1(z2 + z3 ) = z1z2 + z1z3 and (z 2 + z3 )z1 = z2 z1 + z3 z1.
Note : 0 = 0 + 0i is the identity element for addition.
1 = 1 + 0 i is the identity element for multiplication.
The additive inverse of a complex number z = a + ib is − z (i.e. – a – ib).
• For every non-zero complex number z, the multiplicative inverse of z is 1 .
z
Example: 7 1 − 2i + 4−i = [Rajasthan PET 1987]
Solution: (d) 2+i 3 + 2i
(a) 24 + 10 i (b) 24 − 10 i (c) 10 + 24 i (d) 10 − 24 i
13 13 13 13 13 13 13 13
1 − 2i + 4−i = = (1 − 2i)(3 + 2i) + (4 − i)(2 + i) = 50 − 120i = 10 − 24 i.
2+i 3 + 2i (2 + i)(3 + 2i) 65 13 13
Example: 8 1 1 + 3 i 3 + 4i [Roorkee 1979; Rajasthan PET 1999]
Solution: (d) − 2i 1+ 2 − 4i
Example: 9
(a) 1 + 9 i (b) 1 − 9 i (c) 1 − 9 i (d) 1 + 9 i
2 2 2 2 4 4 4 4
1 + 3 3 + 4i = 1 + 2i + 3 − 3i 6 − 16 + 12i + 8i = 2 + 4i + 15 − 15i −1 + 2i
− 2i 1+ 2 − 4i 12 + 22 12 + 12 22 + 42 10 2
1 i
= (17 − 11i)(−1 + 2i) = 5 + 45i = 1 + 9 i.
20 20 4 4
The real value of θ for which the expression 1 + i cosθ is a real number, is
1 − 2i cosθ
(a) nπ + π / 2 (b) nπ − π / 2 (c) nπ ± π / 2 (d) None of these
Solution: (c) Given that 1 + i cosθ = (1 + i cosθ )(1 + 2i cosθ ) = (1 − 2 cos2 θ) + i 3 cos θ
1 − 2i cosθ (1 − 2i cosθ )(1 + 2i cosθ ) 4 cos2 θ ) 1 + 4 cos2
(1 + θ
Since Im (z) = 0 , then 3 cosθ = 0 ⇒ θ = nπ ± π / 2.
Equality of Two Complex Numbers.
Two complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 are said to be equal if and only if their real parts and
imaginary parts are separately equal.
i.e., z1 = z2 ⇒ x1 + iy1 = x2 + iy2 ⇔ x1 = x2 and y1 = y2.
Thus , one complex equation is equivalent to two real equations.
Note : A complex number z = x + iy = 0 iff x = 0, y = 0.
The complex number do not possess the property of order i.e., (a + ib) < (or) > (c + id) is not
defined. For example, the statement 9 + 6i > 3 + 2i makes no sense.
Example: 10 Which of the following is correct
Solution: (d)
Example: 11 (a) 6 + i > 8 − i (b) 6 + i > 4 − i (c) 6 + i > 4 + 2i (d) None of these
Because, inequality is not applicable for a complex number. [MP PET 2000; IIT 1998]
6i − 3i 1
If 4 3i − 1 = x + i y , then
20 3 i
(a) x = 3, y=1 (b) x =1, y=3 (c) x = 0, y=3 (d) x = 0, y=0
Solution: (d) 6i − 3i 1
4 3i − 1 Applying C2 → C2 + 3 i C3
20 3 i
6i 0 1
4 0 − 1 = 0 = 0+ 0 i, Equating real and imaginary parts x = 0, y = 0
20 0 i
Example: 12 The real values of x and y for which the equation (x4 + 2xi) − (3x2 + yi) = (3 − 5i) + (1 + 2yi) is satisfied, are [Roorkee 1984]
(a) x = 2, y = 3 (b) x = −2, y = 1 (c) Both (a) and (b) (d) None of these
3
Solution: (c) Given equation (x4 + 2xi) − (3x2 + yi) = (3 − 5i) + (1 + 2yi) ⇒ (x4 − 3x2) + i(2x − 3y) = 4 − 5i
Equating real and imaginary parts, we get
x4 − 3x2 = 4 .....(i) and 2x − 3y = −5 .....(ii)
Form (i) and (ii), we get x = ±2 and y = 3, 1 .
3
Trick: Put x = 2, y = 3 and then x = −2, y = 1 , we see that they both satisfy the given equation.
3
Conjugate of a Complex Number.
(1) Conjugate complex number : If there exists a complex number z = a + i b, (a ,b) ∈ R, then its conjugate
is defined as z = a − i b . Y
Hence, we have Re(z) = z + z and Im (z) = z−z . Geometrically, the conjugate Imaginary axis
2 2i
P(z)
of z is the reflection or point image of z in the real axis. θ X
O –θ Real axis
(2) Properties of conjugate : If z, z1 and z2 are existing complex numbers, Q(z )
then we have the following results:
(i) (z)= z (ii) z1 + z2 = z1 + z2
(iii) z1 − z2 = z1 − z2 (iv) z1z2 = z1 z2, In general z1.z2.z3.....zn = z1.z2.z3.....zn
(v) z1 = z1 , z2 ≠0 (vi) (z)n = (z n )
z2 z2 (viii) z − z = 2i Im(z) = purely imaginary
(vii) z + z = 2 Re(z) = 2 Re(z) = purely real
(ix) z z = purely real (x) z1z2 + z1z2 = 2Re(z1z2) = 2Re(z1z2)
(xi) z − z = 0 i.e., z = z ⇔ z is purely real i.e., Im(z) = 0
(xii) z + z = 0 i.e., z = −z ⇔ either z = 0 or z is purely imaginary i.e., Re(z) = 0
(xiii) z1 = z2 ⇔ z1 = z2 (xiv) z = 0 ⇔ z = 0
(xv) zz = 0 ⇔ z = 0 (xvi) If w = f(z)then w = f(z) (xvii) reiθ = re−iθ
Important Tips
• Complex conjugate is obtained by just changing the sign of i.
• Conjugate of i = −i
• Conjugate of iz = −iz
• (z1 + z2 ) and (z1 . z2 ) real ⇔ z1 = z2 or z2 = z1
• z1z2 = z1z2
(3) Reciprocal of a complex number : For an existing non-zero complex number z = a + ib , the
reciprocal is given by z −1 = 1 z i.e., z −1 = 1 ⇒ a − ib = Re(z) + i[− Im(z)] = | z .
z = | z |2 a + ib a2 + b2 | z |2 | z |2 z |2
Example: 13 If the conjugate of (x + i y) (1 − 2 i) be 1+i, then [MP PET 1996]
Solution: (c)
Example: 14 (a) x= 1 (b) y= 3 (c) x + i y = 1−i i (d) x −iy = 1−i
Solution: (c) 5 5 1− 2 1 + 2i
Example: 15
Solution: (d) Given that (x + i y) (1− 2i) = 1 + i ⇒ x − i y = 1+i i ⇒x+iy = 1−i
1+ 2 1− 2i
Example: 16
Solution: (b) For the complex number z, one from z + z and z z is [Rajasthan PET 1987]
Example: 17
Solution: (c) (a) A real number (b) An imaginary number
(c) Both are real numbers (d) Both are imaginary numbers
Example: 18
Solution: (c) Here z + z = (x + i y)+ (x − i y) = 2x (Real) and zz =(x + i y)(x − iy)= x 2 + y2 (Real).
The complex numbers sin x + i cos 2x and cos x − i sin 2x are conjugate to each other for [IIT 1988]
(a) x = nπ (b) x = n + 1 π (c) x = 0 (d) No value of x
2
sin x + i cos 2x and cos x − i sin 2x are conjugate to each other if sin x = cos x and cos 2x = sin 2x
or tan x = 1⇒ x = π , 5π , 9π ,........ (i) and tan 2x = 1⇒ 2x = π , 5π , 9π ,........ or x = π , 5π , 9π ,....... (ii) There exists no
4 4 4 4 4 4 8 8 8
value of x common in (i) and (ii). Therefore there is no value of x for which the given complex numbers are conjugate.
The conjugate of complex number 2 − 3i is [MP PET 2004]
4−i
(a) 3i (b) 11 + 10i (c) 11 − 10i (d) 2 + 3i
4 17 17 4i
2 − 3i = (2 − 3i)(4 + i) = 8 + 3 − 12i + 2i = 11 − 10i ⇒ Conjugate = 11 + 10i .
4−i (4 − i)(4 + i) 16 + 1 17 17
The real part of (1 − cos θ + 2i sin θ )−1 is [Karnataka CET 2001; IIT 1978, 86]
(a) 1 (b) 1 (c) 1 (d) 1
3 + 5 cosθ 5 − 3 cosθ 3 − 5 cosθ 5 + 3 cosθ
i. 2 sinθ }−1 = 2 sin2 θ θ θ −1 2 θ −1 sin θ θ −1
2 2 2 2 2
{(1 − cosθ ) + + i.4 sin cos 2 = sin + i. 2 cos
2 sin θ −1. 1 sin θ − i.2 cos θ sin θ − i.2 cos θ
2 + i.2 sin 2 − i.2 cos 2 2 2
= θ θ × θ θ =
2 2 2 2 θ sin2 θ θ
sin cos 2 sin 2 2 + 4 cos2 2
sin θ 1 1
2 5 + 3 cosθ
Hence, real part = = = .
θ 1 θ 21 θ
2 sin 2 + 3 cos 2 2 + 3 cos 2 2
The reciprocal of 3 + 7i is [Kurukshetra CEE 2000]
(a) 3 − 7 i (b) 3 − 7i (c) 3 − 7 i (d) 7 + 3i
4 4 16 16
1 7i = 1 7i . 3− 7i = 3 − 7i = 3 − 7i = 3 − 7 i.
3+ 3+ 3− 7i 9+7 16 16 16
Modulus of a Complex Number.
Modulus of a complex number z = a + ib is defined by a positive real number given by | z |= a 2 + b 2 ,
where a, b real numbers. Geometrically |z| represents the distance of point P (represented by z) from the origin,
i.e. |z| = OP. Y
If |z| = 0, then z is known as zero modular complex number and is P(z)
used to represent the origin of reference plane.
If |z| = 1 the corresponding complex number is known as
unimodular complex number. Clearly z lies on a circle of unit radius O MX
having centre (0, 0).
Note : In the set C of all complex numbers, the order relation is not defined. As such z1 > z2 > or
z1 < z2 has no meaning. But | z1 |>| z2 |or| z1 |<| z2 | has got its meaning since | z1 |and| z2 | are
real numbers.
Properties of modulus
(i) z ≥ 0 ⇒ z = 0 iff z = 0 and |z|> 0 iff z ≠ 0 .
(ii) − z ≤ Re (z) ≤ z and − z ≤ Im (z)≤ z
(iii) z = z = − z = − z =| zi|
(iv) zz = z 2 =| z |2
(v) z1z2 = z1 z2 . In general z1z2z3 ......zn = z1 z2 z3 .... zn
(vi) z1 = z1 , (z 2 ≠ 0)
z2 z2
(vii) | z n |=| z |n,n ∈ N
(viii) z1 ± z2 2= (z1 ± z 2 )(z1 ± z 2 ) = z1 2 z2 2 ± (z1z 2 + z1z 2 ) or | z1 |2 + | z2 |2 ± 2 Re(z1z2)
+
(ix) z1 + z2 2 = z1 2+ z2 2 ⇔ z1 is purely imaginary or Re z1 = 0
z2 z2
{ }(x) z1 + z2 2 + z1 − z2 2 = 2 z1 2 + z2 2 (Law of parallelogram)
( )(xi) az1 − bz2 2 + bz1 + az2 2 = (a 2 + b 2 ) z1 2 + z2 2 , where a,b ∈ R.
Important Tips
• Modulus of every complex number is a non-negative real number. | z |= 0 iff z = 0 i.e., Re(z) = Im(z) = 0
• | z | ≥|Re(z)| ≥ Re(z) and | z |≥|Im(z)|≥ Im(z) | z |= 1⇔ z = 1
z
• z =1 | z | is always a unimodular complex number if z ≠ 0
z z
• z is always a unimodular complex number if z≠0 |Re(z)| + |Im(z)| ≤ 2 | z |
z
• || z1 | −| z2 ||≤| z1 + z2 |≤| z1 | + | z2 |
Thus | z1 | + | z2 | is the greatest possible value of | z1 + z2 | and || z1 | −| z2 || is the least possible value of | z1 + z2 |
• If z + 1 = a, the greatest and least values of | z |are respectively a+ a2 + 4 and −a+ a2 + 4
z 2 2
• | z1 + z12 − z 2 |+| z1 − z12 − z 2 |=| z1 + z2 |+| z1 − z2 |
2 2
Example: 19 (1 + i) (2 + i) = [MP PET 1995, 99; Rajasthan PET 1998]
Solution (c) (3 + i)
(d) – 1
(a) – 1/2 (b) 1/2 (c) 1
z = (1 + i)(2 + i) = 1 + 3i × 3−i = 3 + 4i ⇒ |z|=1
(3 + i) 3+i 3−i 5
Trick : |z|= |z1||z 2 | = 2. 5 = 1
|z3 | 10
Example: 20 If α and β are different complex numbers with |β|=1, then β −α is equal to [IIT 1992]
1−αβ (d) 2
(a) 0 (b) 1/2 (c)
1
Solution (c) β −α = β −α = β −α , = 1 β −α =|β1| = 1 {|z|=|z| }
Example: 21 1 − αβ ββ − αβ β(β − α ) |β| (β −α)
Solution (b)
Example 22 For any complex number z, maximum value of |z|−|z − 1| is
Solution: (b)
Example: 23 (a) 0 (b) 1 (c) 3/2 (d) None of these
Solution: (d) We know that |z1 − z 2 |≥| z1|−|z 2 |
∴|z|−|z − 1|≤|z − (z − 1)|or |z|−|z − 1|≤ 1 , ∴ Maximum value of |z|−|z − 1| is 1.
If z = x + iy and iz 2 − z = 0 , then | z | is equal to [Bihar CEE 1998]
(a) 1 (b) 0 or 1 (c) 1 or 2 (d) 2
iz 2 = z ⇒| iz 2 |=|z|⇒|z|2 =|z| ⇒|z|(|z |− 1) = 0 ⇒ |z|= 0 or | z | = 1
For x1, x2, y1, y2 ∈ R, if 0 < x1 < x 2 , y1 = y2 and z1 = x1 + iy1, z 2 = x 2 + iy2 , and z3 = 1 (z1 + z 2 ), then z1, z2 and z3
2
satisfy [Roorkee 1991]
(a) |z1 |=|z 2 |=| z3 | (b) |z1 |<|z 2 |<| z3 | (c) |z1 |>|z 2 |>| z3 | (d) |z1 |<|z3 |<| z 2 |
0 < x1 < x 2 , y1 = y2 (Given)
|z1 |= x12 + y12 , |z 2 |= x 2 + y 2 ⇒|z 2 |>| z1 | ⇒|z3 |=| z1 + z 2 | = x1 + x2 2 + y1 + y2 2
2 2 2 2 2
x1 + x2 2 + y12 <| z2 |>| z1 |. Hence, |z1 |<|z3 |<| z2 |
2
Argument of a Complex Number.
Let z = a + ib be any complex number. If this complex number is represented geometrically by a point P, then
the angle made by the line OP with real axis is known as argument or amplitude of z and is expressed as
arg (z) =θ = tan −1 b , θ = ∠POM . Also, argument of a complex number Y (+,+)
a P(z)
π–θ
is not unique, since if θ be a value of the argument, so also is 2nπ + θ , where n∈I . (–,+)
π-θ
(1) Principal value of arg (z) : The value θ of the argument, which X' θ θ X
satisfies the inequality − π < θ ≤ π is called the principal value of argument. θ –θ M
Principal values of argument z will be θ ,π − θ ,− π + θ and − θ according as the
(–, –) O (+,–)
point z lies in the 1st , 2nd , 3rd and 4th quadrants respectively, where –π +θ Y' –θ
Y
θ = tan−1 b =α (acute angle). Principal value of argument of any complex π–θ θ
a (–,+)
(+,+)
number lies between − π < θ ≤ π . X' (–,–) O (+,–) X
b – (π – θ) –θ
a
(i) a,b∈ First quadrant a > 0,b > 0 . arg (z) = θ = tan −1 . It is an acute Y'
angle and positive. Y
(a, b)
θ b
Oa
X' X
Y'
(ii) (a,b)∈ Second quadrant, a < 0,b > 0, arg (z) = θ = π − tan −1 |ba| . It is an obtuse angle and positive.
(a, b) y
bθ x
x' a O
y'
(iii) (a, b) ∈ Third quadrant a < 0,b < 0, arg (z) = θ = −π + tan −1 b . It is an obtuse angle and negative.
a
y
x' a O x
bθ
(a, b) y'
Complex numbers
(iv) (a, b)∈ Fourth quadrant a > 0, b < 0, arg (z) = θ = − tan −1 | b | . It is an acute angle and negative.
a
y
x' Oa x
θb
y' (a, b)
Quadrant x y arg(z) Interval of θ
θ
I + + 0<θ <π /2
II – + π −θ π /2<θ <π
III – – −(π −θ ) − π < θ < −π / 2
IV + – −π /2<θ <0
−θ
Note : Argument of the complex number 0 is not defined.
• Principal value of argument of a purely real number is 0 if the real number is positive and is π if
the real number is negative.
• Principal value of argument of a purely imaginary number is π / 2 if the imaginary part is
positive and is − π / 2 if the imaginary part is negative.
(2) Properties of arguments
(i) arg (z1z2) = arg(z1) + arg(z2) + 2kπ , (k = 0 or 1 or – 1)
In general arg (z1z 2 z3 .........zn ) = arg(z1 ) + arg(z 2 ) + arg(z 3 ) + .......... + arg(zn ) + 2kπ , (k = 0 or 1 or − 1)
(ii) arg (z1z2 ) = arg(z1 ) − arg(z2 ) (iii) arg z1 = arg z1 − arg z2 + 2kπ , (k = 0 or 1 or – 1)
z2
(iv) arg z = 2arg z + 2kπ , (k = 0 or 1 or – 1) (v) arg(z n ) = n arg z + 2kπ , (k = 0 or 1 or – 1)
z
(vi) If arg z2 = θ, then arg z1 = 2kπ −θ , where k∈I
z1 z2
(vii) arg z = −arg z = arg 1 (viii) arg (z − z) = ±π / 2
z (x) arg (z) + arg(z) = 0 or arg(z) = −arg(z)
(ix) arg (−z) = arg (z) ± π
(xi) arg(z) − arg(z) = ±π
(xii) z1z 2 + z1z 2 = 2|z1 ||z 2 |cos (θ1 − θ 2 ), where θ1 = arg(z1) and θ2 = arg(z2)
Note : Proper value of k must be chosen so that R.H.S. of (i), (ii), (iii) and (iv) lies in (−π ,π )
The property of argument is same as the property of logarithm.
If arg (z) lies between − π and π (π inclusive), then this value itself is the principal value of arg (z). If
not, see whether arg (z) > π or ≤ − π . If arg(z) > π , go on subtracting 2π until it lies between − π
and π (π inclusive). The value thus obtained will be the principal value of arg (z).
The general value of a rg (z) is 2nπ − arg (z) .
Important Tips
• If z1 = z2 ⇔ z1 = z2 and arg z 1 = arg z2. ( )z1 + z2 2 ≥ z1 + z2 2 − (arg (z1) − arg (z2))2
• z1 + z2 = z1 | +|z2 ⇔ arg (z1) = arg (z 2 ) i.e., z1 and z2 are parallel.
• z1 + z2 = z1 | +|z2 ⇔ arg (z1) − arg (z2) = 2nπ , where n is some integer.
• | z1 − z2 | = || z1 | −| z2 || ⇔ arg(z1) − arg(z2) = 2nπ , where n is some integer.
• z1 + z 2 = z1 − z 2 ⇔ arg (z1 ) – arg (z2) = π / 2 .
( )• If | z1 |≤1,| z2 |≤1 then (i) z1 + z2 2 ≤ z1 − z2 2 + (arg (z1) − arg (z2))2 (ii)
• z1 + z2 2 = z1 2 + z2 2 + 2| z1 || z2 |cos(θ1 − θ2).
• z1 − z2 2 = z1 2 + z2 2 − 2| z1 || z2 |cos(θ1 − θ2).
• If |z1 |=|z 2 | and amp (z1 ) + amp (z 2 ) = 0, then z1 + z 2 are conjugate complex numbers of each other.
• z ≠ 0, amp (z + z) = 0 or π ;amp(zz) = 0;amp(z − z) = ±π / 2.
• arg (1) = 0, arg (−1) = π ; arg (i) = π / 2, arg (−i) = −π / 2.
• arg (z) = π ⇒ Re(z) = Im(z).
4
• Amplitude of complex number in I and II quadrant is always positive and in IIIrd and IVth quadrant is always negative.
• If a complex number multiplied by i (Iota) its amplitude will be increased by π / 2 and will be decreased by π / 2 , if multiplied by –i,
i.e. arg(iz) = π + arg(z) and arg(−iz) = arg(z) − π .
2 2
Complex number Value of argument
0
+ve Re (z)
–ve Re (z) π
+ve Im (z) π /2
–ve Im (z) 3π / 2 or − π / 2
– (z) |θ ± π |, if θ is − ve and + ve respectively
(iz) π + arg (z)
2
–(iz) arg (z) − π
2
(z n ) n. arg (z)
(z1 .z 2 ) arg (z1) + arg (z2)
arg (z1) – arg (z2)
z1
z2
Example: 24 Amplitude of 1 − i is [Rajasthan PET 1996]
1 + i
(a) −π (b) π (c) π (d) π
2 2 4 6
Solution: (a) z = 1 − i × 1 − i =−i
Example: 25 1 + i 1 − i
Solution: (b)
⇒θ = tan −1 −1 = − π (Since z lies on negative imaginary axis)
Example: 26 0 2
Solution: (d)
If z1 = z 2 and amp z1 + amp z 2 = 0 , then [MP PET 1999]
(a) z1 = z 2 (b) z1 = z 2 (c) z1 + z 2 = 0 (d) z1 = z 2
Let z1 = OP, z 2 = OQ Y P
Since amp (z1) = θ ⇒ amp (z 2 ) = –θ
Q is point image of P O θ X
∴ z1 = z2 –θ
Trick : arg z + arg z = 0, ∴z1 must be equal to z 2 . Q
Let z, w be complex numbers such that z + iw = 0 and arg zw = π , then arg z equals (d) π / 4 [AIEEE 2004]
(a) 5π / 4 (b) π / 2 (c) 3π / 4
z + iw = 0
∴ z = iw ⇒ θ + (π / 2 + θ ) = π , ∴ θ = π / 4 .
Example: 27 The amplitude of sin π + i 1 − cos π [Karnataka CET 2003]
Solution: (c) 5 5
(d) π / 15
Example: 28 (a) π / 5 (b) 2π / 5 (c) π / 10
Solution: [c] [MP PET 1987]
sin π + i (1 − cos π ) = 2 sin π . cos π + i 2 sin 2 π = 2 sin π cos π + i sin π
5 5 10 10 10 10 10 10 (d) − 3 + i
For amplitude, tanθ sin π tan π π .
10 10 10
= = ⇒θ =
π
cos 10
If z =4 and arg z= 5π , then z =
6
(a) 2 3 − 2i (b) 2 3 + 2i (c) − 2 3 + 2i
z =4 and arg z = 5π = 150º ,
6
Let z = x + iy , then z =r= x2 + y2 =4 and θ = 5π = 150º
6
∴ x=r cos θ = 4 cos 150º = − 2 3 and y = r sinθ = 4 sin 150º = 4 ⋅ 1 = 2.
2
∴ z = x + i y = −2 3 + 2i.
Trick: Since arg z = 5π = 150º , here the complex number must lie in second quadrant, so (a) and (b) rejected. Also
6
z = 4 , which satisfies (c) only.
Example: 29 If z and ω are to non-zero complex numbers such that | zω |= 1 and arg (z) – arg (ω) = π , then zω is equal to
2
[AIEEE 2003]
(a) 1 (b) – 1 (c) i (d) – i
Solution: (d) | z ||ω |= 1 .....(i) and arg z = π ⇒ z =i ⇒ z =1 .....(ii)
ω 2 ω ω
From equation (i) and (ii),
| z |=|ω |= 1 and z + z = 0; zω + zω = 0 ⇒ zω = −zω = −z ωω ⇒ zω = −i|ω |2 = −i.
ωω ω
Square Root of a Complex Number.
Let a + ib be a complex number such that a + ib = x + iy, where x and y are real numbers. Then
a + ib = x + iy ⇒ a + ib = (x + iy)2 ⇒ a + ib = (x 2 − y2) + 2ixy
⇒ x2 − y2 = a .....(i)
and 2xy = b .....(ii) [On equating real and imaginary parts]
Solving, x = ± a2 + b2 + a and y=± a2 + b2 − a
2 2
∴ a2 + b2 + a − i a2 + b2 − a
a + ib = ± 2
2
Therefore a + ib = | z | +a + i | z |−a for b>0 | z | +a −i | z | −a for b<0.
± 2 = ± 2 2
2
Note : To find the square root of a − ib, replace i by – i in the above results.
The square root of i is ± 1 +i , [Here b = 1]
2
The square root of – i is ± 1 −i , [Here b = –1]
2
Alternative method for finding the square root
(i) If the imaginary part is not even then multiply and divide the given complex number by 2. e.g. z = 8 – 15i
here imaginary part is not even so write z = 1 (16 – 30i) and let a + ib= 16 – 30 i .
2
(ii) Now divide the numerical value of imaginary part of a + i b by 2 and let quotient be P and find all possible
two factors of the number P thus obtained and take that pair in which difference of squares of the numbers is equal
to the real part of a + i b e.g., here numerical value of Im(16 – 30i) is 30. Now 30 = 2 × 15 . All possible way to
express 15 as a product of two are 1×15 , 3× 5 etc. here 5 2 − 3 2 = 16 = Re (16– 30i) so we will take 5, 3.
(iii) Take i with the smaller or the greater factor according as the real part of a + ib is positive or negative and
if real part is zero then take equal factors of P and associate i with any one of them e.g., Re(16 – 30i) > 0, we will
take i with 3. Now complete the square and write down the square root of z.
[ ]e.g., 1 1 1 2 1 (5 − 3i)
z = 2 [ 16 − 30i ]= 2 5 2 + (3i)2 − 2 × 5 × 3i = 2 [ 5 − 3i ] ⇒ z =± 2
Example: 30 The square root of 3 − 4i are [AMU 1992; Kurukshetra CEE 1996; Rajasthan PET 1999]
(a) ± (2 − i) (b) ± (2 + i) (c) ± ( 3 − 2i) (d) ± ( 3 + 2i)
Solution: (a) z = 5 ,∴ 3 − 4i = ± 5 + 3 −i 5−3 =± (2 − i)
Example: 31 2 2
Solution: (a)
2i equals [Roorkee 1989]
(d) None of these
(a) 1 + i (b) 1 − i (c) − 2i
z = 2i = a + b i ⇒ a = 0,b = 2,| z |= 2
∴ z = ± 2+0 +i 2 −0 = ±(1 + i)
2 2
Trick: It is always better to square the options rather than finding the square root.
Representation of Complex Number.
A complex number can be represented in the following from:
(1) Geometrical representation (Cartesian representation): The complex number z = a + ib = (a, b) is
represented by a point P whose coordinates are referred to rectangular axes Imaginary axis P(a, b)
XOX ′ and YOY ′ which are called real and imaginary axis respectively. Thus a
complex number z is represented by a point in a plane, and corresponding to |z|= a2 + b2 b Real
every point in this plane there exists a complex number such a plane is called O axis
argand plane or argand diagram or complex plane or gaussian plane. θ
aM
Note : Distance of any complex number from the origin is called
the modules of complex number and is denoted by |z|, i.e., | z |= a2 + b2
Angle of any complex number with positive direction of x– axis is called amplitude or argument of
z. i.e., amp(z) = arg(z) = tan−1 b
a
(2) Trigonometrical (Polar) representation : In ∆ OPM, let OP = r , then a = r cosθ and b = r sinθ .
Hence z can be expressed as z = r(cos θ + i sin θ )
where r = |z| and θ = principal value of argument of z.
For general values of the argument z = r[cos(2nπ + θ ) + i sin(2nπ + θ )]
Note : Sometimes (cosθ + i sinθ ) is written in short as cisθ .
Complex number 3
(3) Vector representation : If P is the point (a, b) on the argand plane corresponding to the complex
number z = a + ib .
Then OP = aˆi + bˆj , ∴ |OP |= a2 + b2 =| z | and arg z = direction of the vector OP = tan−1 b
a
Therefore, complex number z can also be represented by OP .
(4) Eulerian representation (Exponential form) : Since we have eiθ = cos θ + i sin θ and thus z can be
expressed as z = reiθ , where | z |= r and θ = arg (z)
Note : e−iθ = (cosθ − i sinθ )
eiθ + e−iθ = 2 cos θ , eiθ − e−iθ = 2i sin θ
Example: 32 1 + 7i = [Roorkee 1998]
Solution: (a) (2 − i)2
Example: 33 (a) 2 cos 3π + i sin 3π (b) 2 cos π + i sin π (c) cos 3π + i sin 3π (d) None of these
Solution: (c) 4 4 4 4 4 4
Example: 34
Solution: (a) 1 + 7i = (1 + 7i)(3 + 4i) = −25 + 25i = −1 + i
Example: 35 (2 − i)2 (3 − 4i)(3 + 4i) 25
Solution: (a)
Let z = x + iy = −1 + i , ∴ x = r cosθ = −1 and y = r sinθ = 1
∴ θ = 3π and r = 2 , Thus 1 + 7i = 2 cos 3π + i sin 3π
4 (2 − i)2 4 4
Alternative method: 1 + 7i = 1 + 7i = 2 and arg 1 + 7i = tan−1 7 − tan−1 −4 = tan−1 7 + tan−1 4 = 3π
(2 − i)2 3 − 4i 3 − 4i 3 3 4
∴ 1 + 7i = 2 cos 3π + i sin 3π
(2 − i)2 4 4
If − 1 + − 3 = reiθ , then θ is equal to [Rajasthan PET 1989; MP PMT 1999]
(a) π (b) − π (c) 2π (d) − 2π
3 3 3 3
Here − 1 + − 3 = reiθ ⇒ − 1 + i 3 = reiθ = r cosθ + ir sinθ
Equating real and imaginary part, we get r cosθ = −1 and r sinθ = 3
Hence tanθ = − 3 ⇒ tanθ = tan 2π ⇒ θ = 2π .
3 3
Real part of e eiθ is [Rajasthan PET 1995]
(a) ecosθ [cos(sinθ )] (b) e cosθ [cos(cosθ )] (c) e sinθ [sin(cosθ )] (d) e sinθ [sin(sinθ )]
e eiθ = e(cosθ +i sinθ ) = e cosθ . e i sinθ = e cosθ [cos(i sinθ ) + i sin(sinθ )]
∴ Real part of e eiθ is e cosθ [cos(sinθ )] .
If 1 + x = 2 cos θ , then xn + 1 is equal to [UPSEAT 2001]
x xn
(a) 2 cos nθ (b) 2 sin nθ (c) cos nθ (d) sin nθ
Let x = cosθ + i sinθ = e iθ then xn + 1 = einθ 1 = einθ + e −inθ = cos nθ + i sin nθ + cos nθ − i sin nθ = 2cos nθ .
xn + einθ
Logarithm of a Complex Number.
Let z = x + iy and
log e (x + iy) = a + ib .....(i)
x + iy = r(cosθ + i sinθ ) = reiθ .....(ii)
then x = r cosθ , y = r sinθ , clearly r = x2 + y2 and θ = tan −1 y
x
From equation (ii), log(x + iy) = log e (reiθ ) = log r + log e eiθ = log e r + iθ = log e (x 2 + y2) + i tan −1 y
x
log e (z) = log e | z | + i amp z
Obviously, the general value is Log(z) = log e (z) + 2πni (−π < amp(z) < π )
Example: 36 ii is equal to [EAMCET 1995]
Solution: (b)
Example: 37 (a) eπ / 2 (b) e −π / 2 (c) −π / 2 (d) None of these
Solution: (b) Let A = ii then log A = log ii = i log i ⇒ log A = i log(0 + i) ⇒ log A = i[log1 + iπ / 2] ( (i) = 1 and arg (i) = π / 2)
log A = i[0 + iπ / 2] = −π / 2 ⇒ A = e −π / 2 .
i log x − i is equal to [Rajasthan PET 2000]
x + i
(a) π + 2 tan −1 x (b) π − 2 tan −1 x (c) − π + 2 tan −1 x (d) − π − 2 tan −1 x
Let z i log x − i z log x − i z log x − i x − i log x 2 − 1 − 2ix
x + i i x + i i x + i x − i x2 +1
= ⇒ = ⇒ = × =
⇒ z = log x2 − 1 − i 2x .....(i)
i x2 + 1 x2 + 1
∵ log(a + ib) = log(reiθ ) = log r + iθ = log a2 + b2 + i tan −1 b
a
z x2 − 1 2 + − 2x 2 + i tan−1 − 2x
i x2 + 1 x2 +1 x2 −1
Hence = log (By equation (i))
z = log x4 + 1 − 2x 2 + 4x 2 + i tan −1 1 2x = log 1 + i(2 tan −1 x) = 0 + i(tan−1 x)
i (x 2 + 1)2 − x2
∴ z = i2 2 tan−1 x = −2 tan−1 x = π − 2 tan−1 x.
Geometry of Complex Numbers.
(1) Geometrical representation of algebraic operations on complex numbers
(i) Sum: Let the complex numbers z1 = x1 + iy1 = (x1, y1) and z2 = x2 + iy2 = (x2, y2) be represented by the
points P and Q on the argand plane. Y Q(x2,y2 R(x1+x2, y2+y2)
Then sum of z1 and z2 i.e., z1 + z2 is represented by the point R. Complex )
number z can be represented by OR. y2
= (x1 + x2) + i(y1 + y2) = (x1 + iy1) + (x2 + iy2) = (z1 + z2) = (x1, y1) + (x2, y2)
P(x1,y1)
In vector notation, we have z1 + z2 = OP + OQ = OP + PR = OR
O x2 N y1 M X
L K
(ii) Difference : We first represent − z2 by Q', so that QQ' is bisected at O.
The point R represents the difference z1 − z2. Y Q(x2,y2
In parallelogram ORPQ, OR = QP )
P(x1,y1)
We have in vectorial notation z1 − z2 = OP − OQ = OP + QO O X
= OP + PR = OR = QP . Q'(–x2,–y2) R(x1–x2, y1–y2)
(iii) Product : Let z1 = r1(cosθ1 + i sinθ1) = r1e iθ1
∴ | z1 |= r1 and arg (z1) = θ1 and z2 = r2 (cosθ 2 + i sinθ 2 ) = r2e iθ2 Y
∴ |z2 |= r2 and arg (z2 ) = θ 2 R(z1,z2)
Then, z1z2 = r1r2 (cosθ1 + i sinθ1)(cosθ 2 + i sinθ 2 ) r1r2 r2 Q(z2) P(z1)
= r1r2 {cos(θ1 + θ 2 ) + i sin(θ1 + θ 2 )} θ1 θ2 r1 X
∴ |z1z2 |= r1r2 and arg (z1z2 ) = θ1 + θ 2 O θ1
A
R is the point representing product of complex numbers z1 and z2.
Important Tips
• Multiplication of i : Since z = r (cosθ + i sinθ ) and i = cos π + i sin π then iz = cos π + θ + i sin π + θ
2 2 2 2
Hence, multiplication of z with i then vector for z rotates a right angle in the positive sense.
i.e., To multiply a vector by –1 is to turn it through two right angles.
i.e., To multiply a vector by (cosθ + i sinθ ) is to turn it through the angle θ in the positive sense.
(iv) Division : Let z1 = r1(cosθ1 + i sinθ1) = r1eiθ1 Y P(z1)
∴ | z1 |= r1 and arg (z1) = θ1 Q(z2) X
and z2 = r2(cos θ2 + i sin θ2) = r2eiθ2
∴ | z2 |= r2 and arg (z2) = θ2 r2
θ2 r1 θ1
O θ2 – θ1 θ2
Then z1 r1 (cosθ1 + i sinθ1) (z2 ≠ 0, r2 ≠ 0) R
z2 r2 (cosθ2 + i sinθ2)
=
⇒ z1 = r1 [cos(θ1 − θ 2) + i sin(θ1 − θ 2 )]
z2 r2
∴ z1 = r1 , arg z1 = θ1 −θ2
z2 r2 z2
Note : The vertical angle R is − (θ 2 − θ1) i.e., θ1 − θ 2 .
• If θ1 and θ 2 are the principal values of z1 and z2 then θ1 + θ 2 and θ1 − θ 2 are not necessarily the principal
value of arg (z1z2) and arg (z1 / z2).
Use of Complex Numbers in Co-ordinate Geometry.
(1) Distance formula : The distance between two points P(z1) and Q(z2) is given by
PQ =| z2 − z1 | = |affix of Q – affix of P|
Q(z2)
P(z1)
Note : The distance of point z from origin | z − 0|=| z |=| z − (0 + i 0)|. Thus, modulus of a complex
number z represented by a point in the argand plane is its distance from the origin.
• Three points A(z1), B(z2) and C(z3 ) are collinear then AB + BC = AC
i.e., | z1 − z2 | + | z2 − z3 |=| z1 − z3 |.
(2) Section formula : If R(z) divides the joining of P(z1) and Q(z2) in the ratio m1 : m2(m1,m2 > 0)
(i) If R(z) divides the segment PQ internally in the ratio of m1 : m2 then z = m1z 2 + m2z1
m1 + m2
(ii) If R(z) divides the segment PQ externally in the ratio of m1 : m2 Q(z2)
m2
m1z 2 − m2z1
then z = m1 − m2 m1 R(z) m2
m1
R(z)
Note : If R(z) is the mid point of PQ then affix of R is z1 + z2 P(z1) Q(z2)
2 P(z1)
• If z1, z2, z3 are affixes of the vertices of a triangle, then affix of its centroid is z1 + z2 + z3 .
3
(3) Equation of the perpendicular bisector : If P(z1) and Q(z2) are two fixed points and R(z) is moving
point such that it is always at equal distance from P(z1) and Q(z2) P(z1)
i.e., PR = QR or | z − z1 |=| z − z2 |
⇒ | z − z1 |2 =| z − z2 |2 R(z)
⇒ (z − z1)(z − z1) = (z − z2)(z − z2) Q(z2)
⇒ (z − z1)(z − z1) = (z − z2)(z − z2)
⇒ z z(z1 − z2 ) + z(z1 − z2 ) = z1z1 − z2 z2 ⇒ z(z1 − z 2 ) + z(z1 − z 2 ) =| z1 |2 −| z 2 |2
Hence, z lies on the perpendicular bisector of z1 and z2 .
(4) Equation of a straight line
(i) Parametric form : Equation of a straight line joining the point having affixes z1 and z2 is
z = t z1 + (1 − t)z2, when t ∈ R
(ii) Non parametric form : Equation of a straight line joining the points having affixes z1 and z2 is
z z1
z1 z1 1 = 0 ⇒ z(z1 − z2) − z(z1 − z2) + z1z2 − z2z1 = 0 .
z2 z2 1
Note : z1 z1 1
Three points z1, z2 and z3 are collinear z2 z2 1 = 0
z3 z3 1
(iii) General equation of a straight line: The general equation of a straight line is of the
form az + az + b = 0 , where a is complex number and b is real number.
(iv) Slope of a line : The complex slope of the line az + az + b = 0 is a coeff. of z and real slope of
−a = − coeff. of z
the line az + az +b = 0 is − Re(a) = −i (a + a) .
Im(a) (a − a)
Note : If α1 and α 2 are the are the complex slopes of two lines on the argand plane, then
(i) If lines are perpendicular then α1 + α 2 = 0 (ii) If lines are parallel then α1 = α 2
• If lines az + az + b = 0 and a1z + a1z + b1 = 0 are the perpendicular or parallel, then
−a + − a1 = 0 or −a = − a1 ⇒ aa1 + a1a = 0 or aa1 − aa1 = 0, where a, a1 are the complex
a a1 a a1
numbers and b,b1 ∈ R.
(v) Slope of the line segment joining two points : If A(z1) and B(z2 ) represent two points in the argand
plane then the complex slope of AB is defined by z1 − z2 . B(z2)
z1 − z2
Note : If three points A(z1), B(z2 ), C(z3 ) are collinear then slope of
AB = slope of BC = slope of AC
A(z1)
z1 − z2 z2 − z3 z1 − z3
z1 − z2 = z2 − z3 = z1 − z3
(vi) Length of perpendicular : The length of perpendicular from a point z1 to the line az + az + b = 0 is
given by | az1 + az1 + b | or | az1 + az1 + b |
|a|+|a | 2|a|
(5) Equation of a circle : The equation of a circle whose centre is at point having affix zo and radius r is
| z − zo |= r
Note : If the centre of the circle is at origin and radius r, then its P(z)
equation is | z |= r . r
| z − z0 |< r represents interior of a circle | z − z0 |= r and C(z0)
| z − z0 |> r represent exterior of the circle | z − zo |= r . Similarly,
| z − z0 |> r is the set of all points lying outside the circle and
| z − z0 |≥ r is the set of all points lying outside and on the circle | z − z0 |= r.
(i) General equation of a circle : The general equation of the circle is zz + az + az + b = 0 where a is
complex number and b ∈ R .
∴ Centre and radius are – a and | a |2 −b respectively.
Note : Rule to find the centre and radius of a circle whose equation is given:
• Make the coefficient of zz equal to 1 and right hand side equal to zero.
• The centre of circle will be = – a = −coefficent of z
• Radius = |a|2 − constant term
(ii) Equation of circle through three non-collinear points : Let A(z1 ), B(z 2 ), C(z3 ) are three points on
the circle and P(z) be any point on the circle, then ∠ACB = ∠APB
Using coni method P(z)
In ∆ACB, z2 − z3 = BC e iθ .....(i) θ C(z3)
z1 − z3 CA θ
In ∆ APB, z2 − z BP e iθ .....(ii) A(z1) B(z2)
z1 − z AP
=
From (i) and (ii) we get
(z − z1 )(z 2 − z3) = Real .....(iii)
(z − z2 )(z1 − z3)
(iii) Equation of circle in diametric form : If end points of diameter represented by A(z1 ) and B(z2) and
P(z) be any point on circle then, (z − z1)(z − z2 ) + (z − z2 )(z − z1) = 0
which is required equation of circle in diametric form.
(iv) Other forms of circle : (a) Equation of all circle which are orthogonal to | z − z1 |= r1 and |z − z2 |= r2 .
Let the circle be | z − α |= r cut given circles orthogonally
⇒ r 2 + r12 =|α − z1 |2 …...(i) and r 2 + r22 =|α − z2 |2 …..(ii)
on solving r22 − r12 = α(z1 − z2) + α (z1 − z2)+ | z2 |2 − | z1 |2 and let α = a + ib
P(z)
A(z1) B(z2)
Diameter
(b) z − z1 = k is a circle if k ≠ 1 and a line if k = 1.
z − z2
(c) The equation | z − z1 |2 +|z − z2 |2 = k, will represent a circle if k ≥ 1 | z1 − z2 |2
2
(6) Equation of parabola : Now for parabola SP = PM
| z − a |= | z + z+ 2a | M P(z)
2 N A S(a+i.0)
or zz − 4a(z + z) = 1 {z 2 + (z)2} z+z+2a=0 P(z)
2 S'(z2) S(z1)
where a ∈ R (focus)
Directrix is z + z + 2a = 0
(7) Equation of ellipse : For ellipse SP + S' P = 2a
⇒ | z − z1 | + | z − z2 |= 2a
where 2a >| z1 − z2 | (since eccentricity <1)
Then point z describes an ellipse having foci at z1 and z2 and a ∈ R+ .
(8) Equation of hyperbola : For hyperbola SP − S' P = 2a P(z)
⇒ | z − z1 | − | z − z2 |= 2a
where 2a <| z1 − z2 | (since eccentricity >1)
Then point z describes a hyperbola having foci at z1 and z2 and a ∈ R+ S(z1) O S'(z2)
Example: 38 If in the adjoining diagram, A and B represent complex number z1 and z2 respectively, then C represents
Solution: (a)
(a) z1 + z 2 YC
(b) z1 − z 2 B
(c) z1.z 2
(d) z1 / z 2 A
It is a fundamental concept. OX
Example: 39 If centre of a regular hexagon is at origin and one of the vertex on argand diagram is 1 + 2i, then its perimeter is
Solution: (d)
[Rajasthan PET 1999; Himachal CET 2002]
(a) 2 5 (b) 6 2 (c) 4 5 (d) 6 5
Let the vertices be z0 , z1,....., z5 w.r.t. centre O and | z0 |= 5
⇒ A0 A1 =|z1 − z0 |=|z0 e iθ − z0 | =| z0 || cosθ + i sinθ − 1| = 5 (cosθ − 1)2 + sin 2 θ
⇒ = 5 2(1 − cosθ ) = 5.2 sin(θ / 2) (1+2i)
A4 A3
2π π
⇒ A0 A1 = 5 . 2 sin(π / 6) = 5 θ = 6 = 3 .....(i)
A5 O (0,0) A2
A0 A1
Similarly, A1 A2 = A2 A3 = A3 A4 = A4 A5 = A5 A0 = 5
Hence, the perimeter of regular polygon is
= A0 A1 + A1 A2 + A2 A3 + A3 A4 + A4 A5 + A5 A0 = 6 5 .
Example: 40 The complex numbers z1, z2 and z3 satisfying z1 − z3 = 1−i 3 are the vertices of a triangle which is
Solution: (b) z2 − z3 2
Example: 41
Solution: (c) [IIT Screening2001]
Example: 42 (a) Of area zero (b) Right-angled isosceles (c) Equilateral (d) Obtuse-angled isosceles
Solution: (b)
Example: 43 Taking mod of both sides of given relation z1 − z3 = 1 −i 3 = 1 + 3 =1 .
Solution: (a) z2 − z3 2 2 4 4
So, |z1 − z 3 |=|z 2 − z 3 |. Also, amp z 1 − z3 = tan −1 (− 3) = − π or amp z2 − z3 = π or ∠z 2 z3 z1 = 60°
z 2 − z3 3 z1 − z3 3
∴ The triangle has two sides equal and the angle between the equal sides = 60° . So it is equilateral.
Let the complex numbers z1, z2 and z3 be the vertices of an equilateral triangle. Let z0 be the circumcentre of the
triangle, then z12 + z 2 + z 2 = [IIT 1981]
2 3
(a) z 2 (b) − z 2 (c) 3z 2 (d) − 3z 2
0 0 0 0
Let r be the circum-radius of the equilateral triangle and ω the cube root of unity.
Let ABC be the equilateral triangle with z1, z2 and z3 as its vertices A, B and C Y
A'(z1 – z0) r θ
respectively with circumcentre O'(z0 ). The vectors O' A, O' B, O' C are equal
2π O
and parallel to OA′, OB′, OC' respectively. 3
Then the vectors OA' = z1 − z0 = reiθ 2π X
3 C (z3 – z0)
⇒ OB' = z2 − z0 = rei(θ +2π / 3) = rω eiθ
⇒ OC' = z3 − z0 = rei(θ +4π / 3) = rω 2eiθ
∴ z1 = z0 + reiθ , z2 = z0 + rω eiθ , z3 = z0 + rω 2eiθ A(Z1)
Squaring and adding, we get,
2π
z12 + z22 + z 2 = 3z02 + 2(1 + ω + ω 2)z0reiθ + (1 + ω 2 + ω 4 )r 2ei 2θ = 3z02, 3 O'(Z0)
3
since 1 + ω + ω 2 = 0 = 1 + ω 2 + ω 4 .
The points z1, z2 , z3 , z4 in the complex plane are the vertices of a parallelogram B(Z2) C(Z3)
taken in order, if and only if
[IIT 1981,83]
(a) z1 + z4 = z2 + z3 (b) z1 + z3 = z 2 + z4 (c) z1 + z2 = z3 + z4 (d) None of these
Diagonals of a parallelogram ABCD are bisected each other at a point i.e., z1 + z3 = z2 + z4 ⇒ z1 + z3 = z2 + z4 .
2 2
If the complex number z1, z2 and the origin form an equilateral triangle then z12 + z 2 = [IIT 1983, Karnataka CET 1996]
2
(a) z1z 2 (b) z1z 2 (c) z 2 z1 (d) | z1 |2 =| z 2 |2
Let OA, OB be the sides of an equilateral ∆ OAB and OA, OB represent the complex numbers or vectors
z1, z2 respectively.
From the equilateral ∆ OAB, AB = Z2 − Z1 Y B(Z2)
∴ arg z2 − z1 = arg(z 2 − z1) − argz 2 = π and arg z2 = arg(z2 ) − arg(z1 ) = π Z2 Z2– Z1
z2 3 z1 3
π A(Z1)
z2 − z1 z2 3 Z1 X
z2 z1
Also, =1= , since triangle is equilateral. O
Thus the vectors z2 − z1 and z2 have same modulus and same argument, which implies that the vectors are equal, that
z2 z1
is z2 − z1 = z2 ⇒ z1 z 2 − z12 = z 2 ⇒ z12 + z 2 = z1z2.
z2 z1 2 2
Rotation Theorem.
Rotational theorem i.e., angle between two intersecting lines. This is also known as coni method.
Let z1, z2 and z3 be the affixes of three points A, B and C respectively taken on argand plane.
Then we have AC = z3 − z1 and AB = z2 − z1 Y C(z3)
and let arg AC = arg (z3 − z1) = θ and AB = arg (z2 − z1) = φ
Let ∠CAB = α , ∠CAB = α = θ − φ α B(z2)
A(z1)
= arg AC − arg AB = arg (z3 − z1) − arg (z2 − z1) = arg z3 − z1 φθ X
z2 − z1 O
or angle between AC and AB = arg affix of C − affix of A
affix of B − affix of A
For any complex number z we have z =| z |e i(argz)
Similarly, z 3 − z1 z3 − z1 e ori a rg z3 − z1 z3 − z1 | z 3 − z1 || e i(∠CAB) AC e iα
z 2 − z1 z2 − z1 z2 − z1 z2 − z1 | z 2 − z1 AB
= = =
Note : Here only principal values of the arguments are considered.
• arg z1 − z2 =θ , if AB coincides with CD, then arg z1 − z2 = 0 or ± π , so that z1 − z2 is real.
z3 − z4 z3 − z4 z3 − z4
It follows that if z1 − z2 is real, then the points A, B, C, D P(z1) D
z3 − z4
are collinear. A S(z4) θB
C Q(z2) R(z3)
z1 − z2
If AB is perpendicular to CD, then arg z3 − z4 = ±π / 2,
so z1 − z2 is purely imaginary. It follows that if z1 − z2 = ± k(z3 − z4 ), where k purely imaginary
z3 − z4
number, then AB and CD are perpendicular to each other.
(1) Complex number as a rotating arrow in the argand plane : Let z = r(cosθ + i sinθ ) = reiθ ..…(i)
r. e iθ be a complex number representing a point P in the argand plane. Y Q(zeiφ)
Then OP =| z |= r and ∠POX = θ φ P(z)
Now consider complex number z1 = zeiφ X' θ X
O
or z1 = reiθ .eiφ = rei(θ +φ ) {from (i)}
Clearly the complex number z1 represents a point Q in the argand plane, Y'
when OQ = r and ∠QOX = θ + φ .
Clearly multiplication of z with eiφ rotates the vector OP through angle φ in anticlockwise sense. Similarly
multiplication of z with e−iφ will rotate the vector OP in clockwise sense.
Note : If z1, z2 and z3 are the affixes of the points A,B and C such that AC = AB and
∠CAB = θ . Therefore, AB = z2 − z1, AC = z3 − z1 . C(z3)
B(z2)
Then AC will be obtained by rotating AB through an angle θ in
anticlockwise sense, and therefore,
AC = AB eiθ or (z3 − z1) = (z2 − z1)eiθ or z3 − z1 = eiθ θ
z2 − z1 A(z1)
If A, B and C are three points in argand plane such that AC = AB and ∠CAB = θ then use the
rotation about A to find eiθ , but if AC ≠ AB use coni method.
• Let z1 and z2 be two complex numbers represented by point P and Q in the argand plane such that
∠ POQ = θ . Then, z1eiθ is a vector of magnitude | z1 |= OP along OQ and z1eiθ is a unit vector along
| z1 |
OQ. Consequently, | z2 | . z1eiθ is a vector of magnitude | z2 |= OQ along OQ i.e.,
| z1 |
z2 = | z2 ||.z1e iθ = z2 = z2 .
| z1 z1
(2) Condition for four points to be concyclic : If points A,B,C and D are concyclic ∠ ABD = ∠ ACD
Using rotation theorem A(z1) D(z4)
In ∆ ABD (z1 − z2) = z4 − z2 eiθ .....(i) θ θ
z1 − z2 z4 − z2 .....(ii) B(z2) C(z3)
In ∆ ACD (z1 − z3) = z4 − z3 eiθ
z1 − z3 z4 − z3
From (i) and (ii)
(z1 − z2) (z4 − z3) = (z1 − z2)(z4 − z3) =Real
z1 − z3 z4 − z2 (z1 − z3 )(z4 − z2)
So if z1, z2, z3 and z4 are such that (z1 − z2)(z4 − z3) is real, then these four points are concyclic.
(z1 − z3 )(z4 − z2)
Example: 44 If complex numbers z1, z2 and z3 represent the vertices A, B and C respectively of an isosceles triangle ABC of which
Solution: (d)
∠C is right angle, then correct statement is [Rajasthan PET 1999]
(a) z12 + z 2 + z32 = z1 z 2 z 3 (b) (z3 − z1)2 = z3 − z2
2 (d) (z1 − z2)2 = 2(z1 − z3)(z3 − z2)
(c) (z1 − z2)2 = (z1 − z3 )(z3 − z2) A(z1)
BC = AC and ∠C = π / 2 B(z2) 90°
By rotation about C in anticlockwise sense CB = CAeiπ / 2 C(z3
⇒ (z 2 − z3 ) = (z1 − z3 )e iπ / 2 = i(z1 − z3 ) )
⇒ (z2 − z3)2 = −(z1 − z3)2 ⇒ z22 + z32 − 2z2z3 = −z12 − z32 + 2z1z3
⇒ z12 + z 2 − 2z1z 2 = 2z1z3 + 2z 2 z3 − 2z 2 − 2z1z 2
2 3
⇒ (z1 − z2)2 = 2[(z1 z 3 − z 2 ) − (z1z 2 − z2z3 )] ⇒ (z1 − z2)2 = 2(z1 − z3 )(z3 − z2 ).
3
Example: 45 In the argand diagram, if O,P and Q represents respectively the origin, the complex numbers z and z + iz, then the angle
Solution: (c)
Example: 46 ∠OPQ is [MP PET 2000]
Solution: (a)
(a) π (b) π (c) π (d) 2π
Example: 47 4 3 2 3
It is a fundamental concept.
The centre of a regular polygon of n sides is located at the point z = 0 and one of its vertex z1 is known. If z2 be the
vertex adjacent to z1, then z2 is equal to
(a) z1 cos 2π ± i sin 2π (b) z1 cos π ± i sin π (c) z1 cos π ± i sin π (d) None of these
n n n n 2n 2n
Let A be the vertex with affix z1. There are two possibilities of z2 i.e., z2 can be
obtained by rotating z1 through 2π either in clockwise or in anticlockwise B(z2)
n
2π/n
direction. 2π/n
∴ z2 = z2 i 2π i 2π (| z2 |=| z1 |) A(z1) B(z2)
z1 z1
e2 ⇒ z2 = z1e 2
⇒ z2 = z1 cos 2π ± i sin 2π
n n
Let z1, z2, z3 be three vertices of an equilateral triangle circumscribing the circle | z |= 1 . If z1 = 1 3i and z1, z 2 , z3
2 2+ 2
are in anticlockwise sense then z2 is [Orissa JEE 2002]
(a) 1 + 3i (b) 1 − 3 i (c) 1 (d) – 1
Solution: (d) z2 = z1ei 2π /3 = 1 + 3 i cos 2π + i sin 2π = 1 + 3 i − 1 + 3 i = −3 − 1 = −1.
2 2 3 3 2 2 2 2 4 4
Triangle Inequalities.
In any triangle, sum of any two sides is greater than the third side and difference of any two side is less than
the third side. By applying this basic concept to the set of complex numbers we are having the following results.
(1) | z1 + z2 |≤| z1 |+| z2 | (2) | z1 − z2 |≤| z1 |+| z2 |
(3) | z1 + z2 |≥|| z1 |−| z2 || (4) | z1 − z2 |≥|| z1 |−| z2 ||
Note : In a complex plane | z1 − z2 | is the distance between the points z1 and z2 .
The equality | z1 + z2 |=| z1 |+| z2 | holds only when arg (z1) = arg (z2) i.e., z1 and z2 are parallel.
The equality | z1 − z2 |=|| z1 |−| z2 || holds only when arg (z1) – arg (z2) = π i.e., z1 and z2 are
antiparallel.
In any parallelogram sum of the squares of its sides is equal to the sum of the squares of its
diagonals i.e. | z1 + z2 |2 +| z1 − z2 |2 = 2(| z1 |2 +| z2 |2)
Law of polygon i.e., | z1 + z2 + .... + zn |≤|z1 |+| z2 |+....+| zn |
Important Tips
• The area of the triangle whose vertices are z, iz and z + iz is 1 | z |2 .
2
∑• If z1, z2, z3 be the vertices of a triangle then the area of the triangle is (z2 − z3)| z1 |2 .
4iz1
• Area of the triangle with vertices z, wz and z + wz is 3 | z 2 |.
4
• If z1, z2, z3 be the vertices of an equilateral triangle and zo be the circumcentre, then z12 + z 2 + z 2 + = 3z 2 .
2 3 0
• If z1, z2 , z3 .....zn be the vertices of a regular polygon of n sides and z0 be its centroid, then z12 + z 2 + ..... + z 2 = nz 2 .
2 n 0
• If z1, z2 , z3 be the vertices of a triangle, then the triangle is equilateral iff (z1 − z2 )2 + (z2 − z3 )2 + (z3 − z1)2 = 0 or
z12 + z 2 + z 2 = z1 z 2 + z2z3 + z3z1 or z1 1 + z2 1 + 1 =0.
2 3 − z2 − z3 z3 − z1
• If z1, z2z3 are the vertices of an isosceles triangle, right angled at z2 then z12 + z 2 + z 2 = 2z 2 (z1 + z3 ) .
2 3
• If z1, z2 , z3 are the vertices of right-angled isosceles triangle, then (z1 − z2 )2 = 2(z1 − z3 )(z3 − z2 ) .
• If one of the vertices of the triangle is at the origin i.e., z3 = 0, then the triangle is equilateral iff z12 + z 2 − z1z2 =0.
2
z1 z1′ 1
• If z1, z2 , z3 and z1′ , z′2, z′3 are the vertices of a similar triangle, then z2 z′2 1 = 0 .
z3 z′3 1
• If z1, z2, z3 be the affixes of the vertices A, B, C respectively of a triangle ABC, then its orthocentre is
a(sec A)z1 + b(sec B)z2 + (c sec C)z3 .
a sec A + b sec B + c sec C
Example: 48 The points 1 + 3i, 5 + i and 3 + 2i in the complex plane are [MP PET 1987]
Solution: (b)
Example: 49 (a) Vertices of a right angled triangle (b) Collinear
Solution: (b) (c) Vertices of an obtuse angled triangle (d) Vertices of an equilateral triangle
Example: 50 Let z1 = 1 + 3i, z2 = 5+i and z3 = 3 + 2i . Then area of triangle A = 1 x1 y1 1 = 1 1 31
2 x2 y2 1 2 5
x3 y3 1 3 1 1 = 0 , Hence z1, z2 and
21
z3 are collinear.
If z = x + iy , then area of the triangle whose vertices are points z, iz and z + iz is
[IIT 1986; MP PET 1997, 2001; DCE 1997; AMU 2000; UPSEAT 2002]
(a) 2|z|2 (b) 1 | z |2 (c) | z|2 (d) 3 | z |2
2 2
Let z = x + iy , z + iz = (x − y) + i(x + y) and iz = −y + ix
If A denotes the area of the triangle formed by z, z + iz and iz , then
1 xy 1
2 1
A= x−y x+y 1 (Applying transformation R2 → R2 − R1 − R3 )
−y x
1 xy 1 1 1
2 2 2
We get A = 0 0 −1 = (x2 + y2) = | z |2 .
−y x 0
| z1 + z2 |=| z1 | + | z2 | is possible if [MP PET 1999]
(a) z2 = z1 (b) z2 = 1 (c) arg (z1) = arg(z2) (d) | z1 |=| z2 |
z1
Solution: (c) Squaring both sides, we get
| z1 |2 + | z2 |2 + 2 | z1 || z2 |cos(θ1 − θ 2) =| z1 |2 + | z2 |2 +2| z1 || z2 |
⇒ 2| z1 || z2 |cos(θ1 −θ2) = 2| z1 || z2 | ⇒ cos(θ1 − θ 2) = 1 ⇒ θ1 − θ 2 = 0o ⇒ θ1 = θ 2
Hence arg (z1) = arg (z2)
Trick: Let z1 and z2 are the two sides of a triangle. By applying triangle inequality (z1 + z2 ) is the third side. Equality
holds only when θ1 = θ2 i.e., z1 and z2 are parallel.
Standard Loci in the Argand Plane.
(1) If z is a variable point in the argand plane such that arg (z) = θ , then locus of z is a straight line (excluding
origin) through the origin inclined at an angle θ with x–axis.
(2) If z is a variable point and z1 is a fixed point in the argand plane such that arg (z − z1) = θ , then locus of z
is a straight line passing through the point representing z1 and inclined at an angle θ with x-axis. Note that the
point z1 is excluded from the locus.
(3) If z is a variable point and z1, z2 are two fixed points in the argand plane, then
(i) | z − z1 |=| z − z2 | ⇒ Locus of z is the perpendicular bisector of the line
segment joining z1 and z2
(ii) | z − z1 | + | z − z2 |= constant (≠| z1 − z2 |) ⇒ Locus of z is an ellipse
(iii) | z − z1 |+| z − z2 |=| z1 − z2 | ⇒ Locus of z is the line segment joining z1 and z2
(iv) | z − z1 | −| z − z2 |=| z1 − z2 | ⇒ Locus of z is a straight line joining z1 and z2 but z
does not lie between z1 and z2 .
( )(v) |z − z1|−|z − z2|= constant ≠|z1 − z2| ⇒ Locus of z is a hyperbola.
(vi) | z − z1 |2 +| z − z2 |2 =| z1 − z2 | ⇒ Locus of z is a circle with z1 and z2 as the
extremities of diameter.
(vii) | z − z1 |= k | z − z2 | k ≠ 1 ⇒ Locus of z is a circle.
(viii) arg z − z1 = α (fixed) ⇒ Locus of z is a segment of circle.
z − z2
(ix) arg z − z1 = ±π /2 ⇒ Locus of z is a circle with z1 and z2 as the vertices of
z − z2 diameter.
(x) arg z − z1 = 0 or π ⇒ Locus z is a straight line passing through z1 and z2 .
z − z2
z − z1 z − z1 z z1
z2 − z1 z2 − z1
(xi) The equation of the line joining complex numbers z1 and z2 is given by = or z1 z1 1=0
z2 z2 1
Example: 51 The locus of the points z which satisfy the condition arg z − 1 = π is [Rajasthan PET 2000,2002; MP PET 2001]
z + 1 3
(d) None of these
(a) A straight line (b) A circle (c) A parabola
Solution:(c) We have z − 1 = x + iy − 1 = (x2 + y2 − 1) + 2iy
z + 1 x + iy + 1 (x + 1)2 + y2
⇒ arg z −1 = tan−1 x2 2y − 1
z +1 + y2
Hence tan−1 x2 2y −1 = π
+ y2 3
⇒ x2 2y − 1 = tan π = 3 ⇒ x2 + y2 −1 = 2 y ⇒ x2 + y2 − 2 y−1= 0, which is obviously a circle.
+ y2 3 3 3
Example: 52 If z2 − 1 =| z |2 +1, then z lies on [AIEEE 2004]
(a) An ellipse (b) The imaginary axis (c) A circle (d) The real axis
| z2 − 1|=| z |2 +1
Solution: (b)
⇒ | z − 1|2| z + 1|2 = (zz + 1)2 ⇒ (z − 1)(z − 1)(z + 1)(z + 1) = (zz + 1)2 ⇒ z + z = 0
∴ z lies on imaginary axis.
Example: 53 The locus of the point z satisfying arg z − 1 = k. (where k is non-zero) is [Orissa JEE 2002]
Solution: (a) z + 1
(a) Circle with centre on y–axis (b) Circle with centre on x–axis
(b) A straight line parallel to x–axis (d) A straight line making an angle 60° with x–axis
arg z − 1 = k ⇒ arg (x − 1) + iy = k ⇒ arg[(x − 1) + iy] − arg[(x + 1) + iy] = k
z + 1 (x + 1) + iy
y y
y y x −1 − x +1 y(x + 1) − y(x − 1) 2y
⇒ tan−1 x − 1 − tan−1 x + 1 = k ⇒ tan−1 y2 = k ⇒ tan k = x2 + y2 −1 = x2 + y2 −1
x2 −1
1+
⇒ 2y = x2 + y2 −1 ⇒ x2 + y2 − 2y −1 = 0
tan k tan k
It is an equation of circle whose centre is (−g, − f) = (0, cot k) on y–axis.
Example: 54 The locus of z satisfying the inequality log1 / 3 | z + 1|> log1 / 3 | z − 1| is
Solution: (a)
(a) R(z) < 0 (b) R(z) > 0 (c) I(z) < 0 (d) None of these
log1 / 3 | z + 1|> log1 / 3 | z − 1|
⇒ | z + 1|<| z − 1| ⇒ x2 + 1 + 2x + y2 < x2 + 1 − 2x + y2 ⇒ x < 0 ⇒ Re(z) < 0.
Example: 55 If α + iβ = tan−1(z), z = x + iy and α is constant, the locus of 'z' is [EAMCET 1995; KCET 1996]
(a) x2 + y2 + 2x cot 2α = 1 (b) cot 2α(x2 + y2) = 1 + x (c) x2 + y2 + 2y tan 2α = 1 (d) x2 + y2 + 2x sin 2α = 1
Solution: (a) tan(α + iβ ) = x + iy
∴ tan(α − iβ ) = x − iy (conjugate), α is a constant and β is known to be eliminated
tan 2α = tan(α + iβ + α − iβ ) ⇒ tan 2α = x + iy + x − iy ⇒ 1 − (x 2 + y 2 ) = 2x cot 2α
1 − (x + y2)
2
∴ x2 + y2 + 2x cot 2α = 1.
De' Moivre's Theorem.
(1) If n is any rational number, then (cos θ + i sin θ )n = cos nθ + i sin nθ .
(2) If z = (cosθ1 + i sinθ1)(cosθ 2 + i sinθ 2 )(cosθ 3 + i sinθ 3 ).....(cosθ n + i sinθ n )
then z = cos(θ1 + θ 2 + θ 3 + ..... + θ n ) + i sin(θ1 + θ 2 + θ 3 + ..... + θ n ) , where θ1, θ2,θ3.....θn ∈ R .
(3) If z = r(cosθ + i sinθ ) and n is a positive integer, then z1 / n = r1 / n cos 2kπ + θ + i sin 2kπ + θ ,
n n
where k = 0,1, 2, 3,.....(n − 1) .
(4) If p, q ∈ z and q ≠ 0, then (cosθ + i sinθ )p / q = cos 2kπ + pθ + i sin 2kπ + pθ ,
q q
where k = 0,1, 2, 3.....(q − 1) .
Deductions: If n ∈ Q, then
(i) (cosθ − i sinθ )n = cos nθ − i sin nθ (ii) (cosθ + i sinθ )−n = cos nθ − i sin nθ
(iii) (cosθ − i sinθ )−n = cos nθ + i sin nθ (iv) (sin θ + i cos θ )n = cos n π −θ + i sin n π −θ
2 2
Applications
(i) In finding the expansions of trigonometric functions i.e. cos nθ = cosn θ −nC2 cosn−2 θ sin 2 θ
+nC4 cosn−4 θ sin4 θ – ......
sin nθ = nC1 cosn−1 θ sinθ −nC3 cosn−3 θ sin3 θ +nC5 cosn−5 θ sin5 θ − .......
(ii) In finding the roots of complex numbers.
(iii) In finding the complex solution of algebraic equations.
Note : This theorem is not valid when n is not a rational number or the complex number is
not in the form of cosθ + i sinθ .
Powers of complex numbers : Let z = x + iy = r(cosθ + i sinθ )
∴ z n = r n(cosθ + i sinθ )n = r n(cos nθ + i sin nθ )
Number x + iy form Standard complex form General
1 1+i0 cos 0 + i sin 0 cos 2nπ + i sin 2nπ
–1 – 1+i0 cosπ + i sinπ cos(2n + 1)π + i sin(2n + 1)π
i 0 +i(1) cos π + i sin π cos(4n + 1) π + i sin(4n + 1) π
2 2 2 2
–i 0 +i(–1) cos π − i sin π cos(4n + 1) π − i sin(4n + 1) π
2 2 2 2
1 − i 100
1 + i
Example: 56 If = a + ib , then [MP PET 1998]
(a) a = 2, b = –1 (b) a = 1, b = 0 (c) a = 0, b = 1 (d) a = –1, b = 2
Solution: (b) 1− i × 1− i = −i = cos − π + i sin − π ⇒ (−i)100 = cos(−50π ) + i sin(−50π ) = 1+ i (0) ⇒ a = 1, b = 0
1+ i 1− i 2 2
Example: 57 If xr = cos π + i sin π , then x1. x2.x3...........∞ is
2r 2r
[Rajasthan PET 1990, 2000; Karnataka CET 2000; UPSEAT 1990; Haryana CEE 1998; BIT Ranchi 1996]
(a) –3 (b) –2 (c) –1 (d) 0
Solution: (c) x1. x2. x3..... upto ∞ = cos π + i sin π cos π + i sin π ………………..
2 2 22 22
= cos π + π + ... + i sin π + π + .... = cos π 1 + i sin π 1 = cos π + i sinπ = −1
2 22 2 22 2 2 2 2
1 1
− −
Example: 58 If zr = cos rα + i sin rα , where r = 1,2,3,....., n, then lim z1z2z3.....zn is equal to [UPSEAT 2001]
n2 n2
n→∞
(a) cosα + i sinα (b) cos(α / 2) − i sin(α / 2) (c) eiα / 2 (d) 3 eiα
Solution: (c) zr = cos rα + i sin rα ⇒ z1 = cos α + i sin α ;
n2 n2 n2 n2
z2 = cos 2α + i sin 2α ;………………..
n2 n2
⇒ zn = cos nα + i sin nα ⇒ nli→m∞(z1, z 2 , z3 ,.....zn ) = lim cos α (1 + 2+ 3 + ..... + n) + i sin α (1 + 2+ 3 + ..... + n)
n2 n2 n2 n2
n→∞
= lim cos α n(n + 1) + i sin α n(n + 1) = cos α + i sin α = eiα / 2 .
2n2 2n2 2 2
n→∞
Example: 59 1 + sinθ + i cosθ n = [Kerala (Engg.) 2002]
1 + sinθ − i cosθ
(a) cos nπ − nθ + i sin nπ − nθ (b) cos nπ + nθ + i sin nπ + nθ
2 2 2 2
(c) sin nπ − nθ + i cos nπ − nθ (d) cos n nπ + nθ + i sin n nπ + nθ
2 2 2 2
Solution: (a) 1 + sinθ + i cosθ n = 1 + cos α + i sin α n whereα = π −θ
1 + sinθ − i cosθ 1 + cos α − i sin α 2
2 cos2 α + 2i sin α cos α n cos α + i sin α n cis α n cis α α n = cis(nα)
2 cos2 2 − 2i sin 2 . cos 2 cos 2 2 2 2 2
= α α α = α = = +
2 2 2 2 α α
− i sin 2 cis − 2
= cis n π − θ = cis nπ −θ = cos nπ − nθ + i sin nπ − nθ .
2 2 2 2
Roots of a Complex Number.
(1) nth roots of complex number (z1/n) : Let z = r(cos+ i sinθ ) be a complex number. To find the roots of a
complex number, first we express it in polar form with the general value of its amplitude and use the De' moivre's
theorem. By using De'moivre's theorem nth roots having n distinct values of such a complex number are given by
z1/n = r 1/ n cos 2mπ + θ + i sin 2mπ + θ , where m = 0,1, 2,....., (n − 1).
n n
Properties of the roots of z1/n :
(i) All roots of z1/n are in geometrical progression with common ratio e 2π i /n.
(ii) Sum of all roots of z1/n is always equal to zero.
(iii) Product of all roots of z1/n = (−1)n−1 z.
(iv) Modulus of all roots of z1/n are equal and each equal to r 1/ n or | z |1/ n .
(v) Amplitude of all the roots of z1/n are in A.P. with common difference 2π .
n
(vi) All roots of z1/n lies on the circumference of a circle whose centre is origin and radius equal to | z |1/n . Also
these roots divides the circle into n equal parts and forms a polygon of n sides.
(2) The nth roots of unity : The nth roots of unity are given by the solution set of the equation
x n = 1 = cos 0 + i sin 0 = cos 2kπ + i sin 2kπ
x = [cos 2kπ + i sin 2kπ ]1/n
x = cos 2kπ + i sin 2kπ , where k = 0,1, 2,.....,(n − 1).
n n
Properties of nth roots of unity
(i) Let α = cos 2π + i sin 2π = e i(2π / n), the nth roots of unity can be expressed in the form of a series
n n
i.e.,1,α,α 2,.....α n−1. Clearly the series is G.P. with common difference α i.e., e i(2π / n).
(ii) The sum of all n roots of unity is zero i.e., 1 + α + α 2 + ..... + α n−1 = 0.
(iii) Product of all n roots of unity is (−1)n−1.
(iv) Sum of pth power of n roots of unity
1+α p +α 2p + ..... + α (n−1)p = 0, when p is not multiple of n
n, when p is a multiple of n
(v) The n, nth roots of unity if represented on a complex plane locate their positions at the vertices of a regular
plane polygon of n sides inscribed in a unit circle having centre at origin, one vertex on positive real axis.
Note : xn − 1 = (x − 1)(xn−1 + xn−2 + ..... + x + 1)
(sinθ + i cosθ ) = −i2 sinθ + i cosθ = i(cosθ − i sinθ )
(3) Cube roots of unity : Cube roots of unity are the solution set of the equation x 3 − 1 = 0 ⇒ x = (1)1/ 3
⇒ x = (cos 0 + i sin 0)1/ 3 ⇒ x = cos 2kπ + i sin 2kπ , where k = 0,1,2
3 3
Therefore roots are 1, cos 2π + i sin 2π , cos 4π + i sin 4π or 1, e 2π i / 3 , e 4π i / 3 .
3 3 3 3
Alternative : x = (1)1/ 3 ⇒ x 3 − 1 = 0 ⇒ (x − 1)(x 2 + x + 1) = 0
x = 1, − 1 +i 3 , − 1 −i 3
2 2
If one of the complex roots isω, then other root will be ω 2 or vice-versa.
Properties of cube roots of unity
(i) 1 + ω + ω 2 = 0
(ii) ω 3 = 1
(iii) 1+ωr + ω 2r = 0, if r not a multiple of 3
3, if r is a multiple of 3
(iv) ω = ω 2 and (ω )2 = ω and ω .ω = ω 3 .
(v) Cube roots of unity from a G.P.
(vi) Imaginary cube roots of unity are square of each other i.e., (ω)2 = ω 2 and (ω 2)2 = ω 3.ω = ω .
(vii) Imaginary cube roots of unity are reciprocal to each other i.e., 1 =ω2 and 1 =ω.
ω ω2
(viii) The cube roots of unity by, when represented on complex plane, lie on vertices of an equilateral triangle
inscribed in a unit circle having centre at origin, one vertex being on positive real axis.
(ix) A complex number a + ib, for which |a : b|=1 : 3 or 3 : 1, can always be expressed in terms of i,ω,ω 2.
Note : If ω = −1 +i 3 = e 2π i / 3 , then ω2 = −1−i 3 = e −4π i / 3 = e −2π i / 3 or vice-versa
2 2
ω .ω = ω 3.
a + bω + cω 2 = 0 ⇒ a = b = c, if a, b, c are real.
Cube root of – 1 are − 1,−ω,−ω 2 .
Important Tips ( ) x 2 − x + 1 = (x + ω ) x + ω 2
( ) x 2 − xy + y2 = (x + yω ) x + yω 2
• x 2 + x + 1 = (x − ω)(x − ω 2)
( )• x 2 + xy + y2 = (x − yω ) x − yω 2
• x 2 + y 2 = (x + iy)(x − iy) x3 + y3 = (x + y)(x + yω)(x + yω 2)
• x 3 − y 3 = (x − y)(x − yω)(x − yω 2 ) x 2 + y 2 + z 2 − xy − yz − zx = (x + yω + zω 2 )(x + yω 2 + zω)
• x 3 + y 3 + z 3 − 3xyz = (x + y + z)(x + ω y + ω 2 z)(x + ω 2y + ω z)
Fourth roots of unity : The four, fourth roots of unity are given by the solution set of the equation
x 4 − 1 = 0. ⇒ (x 2 − 1)(x 2 + 1) = 0 ⇒ x = ± 1, ± i
Note : Sum of roots = 0 and product of roots =–1.
• Fourth roots of unity are vertices of a square which lies on coordinate axes.
Continued product of the roots
If z = r(cosθ + i sinθ ) i.e., | z |= r and amp (z) = θ then continued product of roots of z1/ n is
r(cosφ + i sinφ) , where n−1 2mπ + θ = (n − 1)π +θ .
∑= φ = m=0 n
Thus continued product of roots of z1/n = r[cos{(n − 1)π + θ } + i sin{(n − 1)π + θ }] = z, if n is odd
− z, if n is even
Similarly, the continued product of values of z m / n is = zm, if n is odd
(-z)m, if n is even
Important Tips
• If x+ 1 = 2 cosθ or x − 1 = 2i sinθ then x = cos θ + i sinθ , 1 = cos θ − i sinθ , xn + 1 = 2 cos nθ , xn − 1 = 2i sin nθ .
x x x xn xn
• If n be a positive integer then , (1 + i)n + (1 − i)n = 2 n +1 cos nπ .
2 4
• If z is a complex number, then e z is periodic.
• nth root of –1 are the solution of the equation z n + 1 = 0
z n − 1 = (z − 1)(z − α)(z − α 2 ).....(z − α n−1), where α = nth root of unity
∏(n−2)/ 2 (z 2 − 2z cos 2rπ + 1), if n is even.
n
z n − 1 = (z − 1)(z + 1)
r =1
∏ ∏zn ((zn−+r=21)0/)2(n-3z)2/2−z22 z−c2ozsco(2sr + 1)π + 1, if n is even.
n n is odd.
+ 1 =
(2r + 1)π
n + 1, if
r=0
• If x = cosα + i sinα, y = cos β + i sin β, z = cos γ + i sin γ and given, x + y + z = 0, then
(i) 1 + 1 + 1 = 0 (ii) yz + zx + xy = 0 (iii) x2 + y2 + z2 =0 (iv) x3 + y3 + z3 = 3xyz
x y z
then, putting, values if x, y, z in these results
x+y+z = 0⇒ cos α + cos β + cos γ = 0 = sin α + sin β + sin γ ⇒ yz + zx + xy = 0⇒ cos(β + γ ) + cos(γ + α) + cos(α + β ) = 0
sin(β + γ ) + sin(γ + α) + sin(α + β ) = 0
∑x 2 cos 2α = 0
the summation consists 3 terms
+ y2 + z2 =0 ⇒
∑ sin 2α = 0,
x 3 + y 3 + z 3 = 3xyz, gives similarly
∑ ∑cos 3α = 3 cos(α + β + γ ) ⇒ sin 3α = 3 sin(α + β + γ )
∑If the condition given be x + y + z = xyz, then cosα = cos(α + β + γ ) etc.
Example: 60 If the cube roots of unity be 1,ω,ω 2 , then the roots of the equation (x − 1)3 + 8 = 0 are
Solution: (c)
[DCE 2000; IIT 1979; UPSEAT 1986]
(a) − 1,1 + 2ω,1 + 2ω 2 (b) − 1,1 − 2ω,1 − 2ω 2 (c) −1,1,−1 (d) None of these
(x − 1)3 = −8 ⇒ x − 1 = (−8)1/ 3 ⇒ x − 1 = −2,−2ω,−2ω 2 ⇒ x = −1,1 − 2ω,1 − 2ω 2
Example: 61 ω is an imaginary cube root of unity. If (1 + ω 2 )m = (1 + ω 4 )m, then least positive integral value of m is
Solution: (d)
Example: 62 [IIT Screening 2004]
Solution: (c)
Example: 63 (a) 6 (b) 5 (c) 4 (d) 3
Solution: (c) The given equation reduces to (−ω)m = (−ω 2 )m ⇒ ω m = 1 ⇒ m = 3 .
Example: 64 If ω is the cube root of unity, then (3 + 5ω + 3ω 2 )2 + (3 + 3ω + 5ω 2 )2 = [MP PET 1999]
Solution: (a) (a) 4 (b) 0 (c) – 4 (d) None of these
Example: 65 (3 + 5ω + 3ω 2 )2 + (3 + 3ω + 5ω 2 )2 = (3 + 3ω + 3ω 2 + 2ω)2 +(3 + 3ω + 3ω 2 + 2ω 2)2
Solution: (b)
= (2ω )2 + (2ω 2 )2 = 4ω 2 + 4ω 4 = 4(−1) = − 4 (1 + ω + ω 2 = 0,ω 3 = 1)
If i = − 1 , then 4 + 5 − 1 + i 3 334 + 3 − 1 + i 3 365 is equal to [IIT 1999]
2 2 2 2
(a) 1 − i 3 (b) − 1 + i 3 (c) i 3 (d) − i 3
Given equation is 4 + 5 − 1 + i 3 334 + 3 − 1 +i 3 365
2 2 2 2
= 4 + 5ω 334 + 3ω 365 = 4 + 5ω + 3ω 2 =1 + 2ω = 1 + 2 − 1 +i 3 =i 3
2
Let ω is an imaginary cube root of unity then the value of
2(ω + 1)(ω 2 + 1) + 3(2ω + 1)(2ω 2 + 1) + ..... + (n + 1)(nω + 1)(nω 2 + 1) is [Orissa JEE 2002]
(a) n(n + 1) 2 + n (b) n(n + 1) 2 (c) n(n + 1) 2 − n (d) None of these
2 2 2
n
∑2(ω + 1)(ω 2 + 1) + 3(2ω + 1)(2ω 2 + 1) + ..... + (n + 1)(nω + 1)(nω 2 + 1) = (r + 1)(rω + 1)(rω 2 + 1)
r =1
n nn n(n + 1) 2
2
= (r + 1)(r 2ω 3 + rω +rω 2 + 1) = (r + 1)(r 2 − r + 1) = (r 3 − r 2 + r + r 2 − r + 1)
∑ ∑ ∑ ∑ ∑r=1 r=1 r=1
= n n + n.
(r 3 ) + (1) =
r =1 r =1
The roots of the equation x 4 − 1 = 0 , are [MP PET 1986]
(a) 1,1,i,−i (b) 1,−1,i,−i (c) 1,−1,ω,ω 2 (d) None of these
Given equation x4 − 1 = 0 ⇒ (x2 − 1)(x2 + 1) = 0 ⇒ x 2 = 1 and x 2 = −1 ⇒ x = ±1,±i
Shifting the Origin in Case of Complex Numbers. YY
Let O be the origin and P be a point with affix z0. Let a point Q has affix z
Q
with respect to the co-ordinate system passing through O.
P (z0) X
When origin is shifted to the point P(z0 )then the new affix Z of the point Q OM X
with respect to new origin P is given by Z = z − z0 i.e., to shift the origin at z0 we
should replace z by Z + z0 .
Example: 66 If z1, z2, z3 are the vertices of an equilateral triangle with z0 as its circumcentre then changing origin to z0 , then (where
Solution: (a) z1, z2, z3 are new complex numbers of the vertices)
(a) z12 + z22 + z32 = 0 (b) z1z2 + z2z3 + z3z1 = 0 (c) Both (a) and (b) (d) None of these
In an equilateral triangle the circumcentre and the centroid are the same point. So,
z0 = z1 + z2 + z3 ⇒ z1 + z2 + z3 = 3z0 ..... (i)
3
To shift the origin at z0 , we have to replace z1, z2 , z3 and z0 by z1 + z0, z2 + z0, z3 + z0 and 0 + z0 then equation (i)
becomes (z1 + z0) + (z2 + z0) + (z3 + z0) = 3(0 + z0) ⇒ z1 + z2 + z3 = 0
On squaring z12 + z22 + z32 + 2(z1z2 + z2z3 + z3z1) = 0 ..... (ii)
But triangle with vertices z1, z2 and z3 is equilateral, then z12 + z22 + z32 = z1z2 + z2z3 + z3z1 .....(iii)
From (ii) and (iii) we get, 3 (z12 + z22 + z32) = 0 . Therefore, z12 + z22 + z32 = 0.
Inverse Points.
(1) Inverse points with respect to a line : Two points P and Q are said to be the inverse points with
respect to the line RS. If Q is the image of P in RS, i.e., if the line RS is the right bisector of PQ.
P
RS
Q
(2) Inverse points with respect to a circle : If C is the centre of the circle and P,Q are the inverse points
with respect to the circle then three points C,P,Q are collinear, and also CP . CQ = r 2, where r is the radius of the
circle.
CP Q
Example: 67 z1, z2, are the inverse points with respect to the line z a + a z = b if
Solution: (b)
(a) z1a + z2a = b (b) z1a + a z2 = b (c) z1a − a z2 = b (d) None of these
Let RS be the line represented by the equation z a + a z = b .....(i) P ≡ z1
Let P and Q are the inverse points with respect to the line RS. R A(z) S
Q ≡ z2
The point Q is the reflection (inverse) of the point P in the line RS if the line
RS is the right bisector of PQ. Take any point z in the line RS, then lines
joining z to P and z to Q are equal.
i.e., | z − z1 |=| z − z2 | or | z − z1 |2 =| z − z2 |2
i.e., (z − z1)(z − z1) = (z − z2 )(z − z2 ) ⇒ z(z2 − z1) + z(z2 − z1) + (z1 z1 − z2 z2 ) = 0 .....(ii)
Hence, equations (i) and (ii) are identical, therefore comparing coefficients, we get
a = a = −b So that, z1a = az2 = −b = z1a + az2 −b
z2 − z1 z2 − z1 z1 z1 − z2 z2 z1(z 2 − z1 ) z 2 (z 2 − z1 ) z1 z1 − z2 z2 0
(By ratio and proportion rule)
Hence, z1a + az2 − b = 0 or z1a + az2 = b.
Example: 68 Inverse of a point a with respect to the circle | z − c |= R (a and c are complex numbers, centre C and radius R) is the point
Solution: (a)
c + R2
a −c
(a) c + R2 (b) c − R2 (c) c + c R (d) None of these
a −c a −c −a
Let a' be the inverse point of a with respect to the circle | z − c |= R, then by definition the points c, a, a' are collinear.
We have, arg(a'−c) = arg(a − c) = −arg(a − c) ( argz = −argz)
⇒ arg(a'−c) + arg(a − c) = 0 ⇒ arg{(a'−c)(a − c)} = 0
∴ (a'−c)(a − c) is purely real and positive. Ca a'
|z – c| = R
By definition |a'−c ||a − c |= R 2 ( CP.CQ = r 2 )
⇒ |a'−c ||a − c |= R 2 (| z |=| z |)
⇒ |(a'−c)(a − c)|= R 2 ⇒ (a'−c)(a − c) = R 2 { (a'−c)(a − c) is purely real and positive}
⇒ a'−c = R2 . Therefore, the inverse point a' of a point a, a' = c + R2 .
a −c a −c
Dot and Cross Product.
Let z1 = a1 + ib1 ≡ (a1,b1) and z2 = a2 + ib2 ≡ (a2 ,b2 ) be two complex numbers.
If ∠ POQ = θ then from coni method z2 −0 = | z2 | e iθ Z2(a2, b2)
z1 −0 | z1 | Z1(a1, b1)
⇒ z 2 z1 = |z2 | e iθ ⇒|zz21z|12 = | z2 | e iθ ⇒ z2z1 =| z1 || z2 | eiθ θ (From(i))
z1z1 | z1 | | z1 | O (From(ii))
⇒ z2z1 =| z1 || z2 |(cosθ + i sinθ )
⇒ Re(z2z1) =| z1 || z2 |cosθ .....(i) and Im(z2z1) =| z1 || z2 |sinθ .....(ii)
The dot product z1 and z2 is defined by z1o z2 =| z1 || z2 |cosθ = Re(z1z2 ) = a1a2 + b1b2
Cross product of z1 and z2 is defined by z1 × z2 =| z1 || z2 |sinθ = Im(z1z2 ) = a1b2 − a2b1
Hence, z1oz 2 = a1a2 + b1b2 = Re(z1z2 ) and z1 × z2 = a1b2 − a2b1 = Im(z1z2)
Important Tips
• If z1 and z2 are perpendicular then z1o z2 = 0 If z1 and z2 are parallel then z1 × z2 = 0
• Projection of z1 on z2 = (z1o z2 ) / | z2 | Projection of z2 on z1 = (z1o z2 ) / | z1 |
• Area of triangle if two sides represented by z1 and z 2 is 1 | z1 × z 2 | Area of a parallelogram having sides z1 and z2 is | z1 × z2 |
2
• Area of parallelogram if diagonals represents by z1 and z2 is 1 | z1 × z2 |
2
Example: 69 If z1 = 2 + 5i, z2 = 3 − i then projection of z1 on z2 is
Solution: (b)
(a) 1/10 (b) 1 / 10 (c) −7 / 10 (d) None of these
Projection of z1 on z2 = z1o z2 = a1a2 + b1b2 = 1.
|z2 | a 2 b22 10
2 +
Permutations-&-combinations
Permutations
The Factorial.
Factorial notation: Let n be a positive integer. Then, the continued product of first n natural numbers is
called factorial n, to be denoted by n ! or n . Also, we define 0 ! = 1.
when n is negative or a fraction, n ! is not defined.
Thus, n ! = n (n – 1) (n – 2) ......3.2.1.
Deduction: n ! = n(n – 1) (n – 2) (n – 3) ......3.2.1
= n[(n − 1)(n − 2)(n − 3)......3.2.1] = n[(n − 1)!]
Thus, 5! = 5 × (4!), 3! = 3 × (2!) and 2! = 2 × (1!)
Also, 1! = 1 × (0!) ⇒ 0! = 1 .
Exponent of Prime p in n ! .
Let p be a prime number and n be a positive integer. Then the last integer amongst 1, 2, 3, .......(n – 1), n
which is divisible by p is n p , where n denote the greatest integer less than or equal to n .
p p p
For example: 10 = 3, 12 = 2, 15 = 5 etc.
3 5 3
Let Ep(n) denotes the exponent of the prime p in the positive integer n. Then,
E p (n!) = E p (1.2.3..........(n − 1)n)= Ep p.2p.3 p....... n p = n + Ep 1.2.3...... n
p p p
[ Remaining integers between 1 and n are not divisible by p]
Now the last integer amongst 1, 2, 3,..... n which is divisible by p is
p
n / p = n = n + Ep p, 2p, 3 p.... n p because the remaining natural numbers from 1 to n are not
p2 p2 p
p p
divisible by p = n + n + E 1.2.3...... n
p p2
p 2 p
Similarly we get E p (n!) = n + n + n + ..... n
p pS
p 2 p 3
where S is the largest natural number. Such that p S ≤ n < p S+1 .
Fundamental Principles of Counting .
(1) Addition principle : Suppose that A and B are two disjoint events (mutually exclusive); that is, they
never occur together. Further suppose that A occurs in m ways and B in n ways. Then A or B can occur in m + n
ways. This rule can also be applied to more than two mutually exclusive events.
Example: 1 A college offers 7 courses in the morning and 5 in the evening. The number of ways a student can select exactly one
course, either in the morning or in the evening
(a) 27 (b) 15 (c) 12 (d) 35
Solution: (c) The student has seven choices from the morning courses out of which he can select one course in 7 ways.
For the evening course, he has 5 choices out of which he can select one course in 5 ways.
Hence he has total number of 7 + 5 = 12 choices.
(2) Multiplication principle : Suppose that an event X can be decomposed into two stages A and B. Let
stage A occur in m ways and suppose that these stages are unrelated, in the sense that stage B occurs in n ways
regardless of the outcome of stage A. Then event X occur in mn ways. This rule is applicable even if event X can be
decomposed in more than two stages.
Note : The above principle can be extended for any finite number of operations and may be stated as under :
If one operation can be performed independently in m different ways and if second operation can
be performed independently in n different ways and a third operation can be performed
independently in p different ways and so on, then the total number of ways in which all the
operations can be performed in the stated order is (m × n × p × .....)
Example: 2 In a monthly test, the teacher decides that there will be three questions, one from each of exercise 7, 8 and 9 of the text
Solution: (a)
book. If there are 12 questions in exercise 7, 18 in exercise 8 and 9 in exercise 9, in how many ways can three questions
be selected
(a) 1944 (b) 1499 (c) 4991 (d) None of these
There are 12 questions in exercise 7. So, one question from exercise 7 can be selected in 12 ways. Exercise 8 contains 18
questions. So, second question can be selected in 18 ways. There are 9 questions in exercise 9. So, third question can be
selected in 9 ways. Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways.
Definition of Permutation.
The ways of arranging or selecting a smaller or an equal number of persons or objects at a time from a given
group of persons or objects with due regard being paid to the order of arrangement or selection are called the
(different) permutations.
For example : Three different things a, b and c are given, then different arrangements which can be made by
taking two things from three given things are ab, ac, bc, ba, ca, cb.
Therefore the number of permutations will be 6.
Number of Permutations without Repetition.
(1) Arranging n objects, taken r at a time equivalent to filling r places from n things
r-places : 12 34 r
Number of choices : n (n–1)(n – 2)(n – 3) n – (r–1)
The number of ways of arranging = The number of ways of filling r places.
= n(n − 1)(n − 2).......(n − r + 1) = n(n − 1)(n − 2).....(n − r + 1)((n − r)!) = n! =n Pr
(n − r)! − r)!
(n
(2) The number of arrangements of n different objects taken all at a time = nPn = n!
Note : n P0 = n! = 1; n Pr = n.n−1 Pr−1
n!
0! = 1; 1 = 0 or (−r)! = ∞ (r ∈ N)
(−r)!
Example: 3 If n P4 : n P5 = 1 : 2 , then n = [MP PET 1987; Rajasthan PET 1996]
(a) 4 (b) 5 (c) 6 (d) 7
Solution: (c) n P4 = 1 ⇒ n! × (n − 5)! = 1 ⇒ n−4 = 2 ⇒ n=6.
Example: 4 n P5 2 (n − 4)! n! 2
Solution: (c)
Example: 5 In a train 5 seats are vacant then how many ways can three passengers sit [Rajasthan PET 1985; MP PET 2003]
Solution: (a)
Example: 6 (a) 20 (b) 30 (c) 60 (d) 10
Solution: (a) Number of ways are = 5 P3 = 5! = 5! = 120 = 60 .
(5 − 3)! 2! 2
Example: 7
How many words comprising of any three letters of the word “UNIVERSAL” can be formed
Solution: (c)
(a) 504 (b) 405 (c) 540 (d) 450
Required numbers of words = 9 P3 = 9! = 9! = 504 .
(9 − 3)! 6!
How many numbers of five digits can be formed from the numbers 2, 0, 4, 3, 8 when repetition of digit is not allowed
[MP PET 2000]
(a) 96 (b) 120 (c) 144 (d) 14
Given numbers are 2, 0, 4, 3, 8
Numbers can be formed = {Total – Those beginning with 0}
= {5 ! – 4 !} = 120 – 24 = 96.
How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 3000 (repetition is not
allowed) [IIT 1976]
(a) 180 (b) 360 (c) 1380 (d) 1500
All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore number of 5 digit numbers
= 6 P5 −5 P5 = 600 .
{Since the case that 0 will be at ten thousand place should be omit}. Similarly number of 6 digit numbers 6 ! – 5 ! = 600.
Now the numbers of 4 digit numbers which are greater than 3000, having 3, 4 or 5 at first place, this can be done in 3
ways and remaining 3 digit may be filled from remaining 5 digits i.e., required number of 4 digit numbers are
5 P3 × 3 = 180 .
Hence total required number of numbers = 600 + 600 + 180 = 1380.
Number of Permutations with Repetition.
(1) The number of permutations (arrangements) of n different objects, taken r at a time, when each object
may occur once, twice, thrice,........upto r times in any arrangement = The number of ways of filling r places where
each place can be filled by any one of n objects.
r – places : 12 34 r
Number of choices : nn n n n
The number of permutations = The number of ways of filling r places = (n)r
(2) The number of arrangements that can be formed using n objects out of which p are identical (and of one
kind) q are identical (and of another kind), r are identical (and of another kind) and the rest are distinct is n! r! .
p!q!
Example: 8 The number of arrangement of the letters of the word “CALCUTTA” [MP PET 1984]
Solution: (b)
(a) 2520 (b) 5040 (c) 10080 (d) 40320
Required number of ways = 8! = 5040 . [since here 2C’s, 2T’s and 2A’s]
2 ! 2! 2!
Example: 9 The number of 5 digit telephone numbers having at least one of their digits repeated is [Pb. CET 2000]
Solution: (d)
(a) 90,000 (b) 100,000 (c) 30,240 (d) 69,760
Using the digits 0, 1, 2,.......,9 the number of five digit telephone numbers which can be formed is 105 .
(since repetition is allowed)
The number of five digit telephone numbers which have none of the digits repeated = 10 P5 = 30240
∴ The required number of telephone numbers = 105 − 30240 = 69760 .
Example: 10 How many words can be made from the letters of the word ‘COMMITTEE’ [MP PET 2002; RPET 1986]
Solution: (b)
(a) 9! (b) 9! (c) 9! (d) 9 !
(2!)2 (2!)3 2!
Number of words = 9! 9! [Since here total number of letters is 9 and 2M’s, 2T’s and 2E’s]
2!2!2! = (2!)3
Conditional Permutations.
(1) Number of permutations of n dissimilar things taken r at a time when p particular things always occur
= n−p Cr−p r!
(2) Number of permutations of n dissimilar things taken r at a time when p particular things never occur
= n−p Cr r!
(3) The total number of permutations of n different things taken not more than r at a time, when each thing
may be repeated any number of times, is n(nr − 1) .
n−1
(4) Number of permutations of n different things, taken all at a time, when m specified things always come
together is m!× (n − m + 1)!
(5) Number of permutations of n different things, taken all at a time, when m specified things never come
together is n!−m! × (n − m + 1)!
(6) Let there be n objects, of which m objects are alike of one kind, and the remaining (n − m) objects are
alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed from
these objects is (m ! ) n! − m) ! .
× (n
Note : The above theorem can be extended further i.e., if there are n objects, of which p1 are alike of
one kind; p2 are alike of another kind; p3 are alike of 3rd kind;......: pr are alike of rth kind such that
p1 + p2 + ...... + pr = n; then the number of permutations of these n objects is (p1 !) × (p n! ...... × (pr ! ) .
!)×
2
Important Tips
• Gap method : Suppose 5 males A, B, C, D, E are arranged in a row as × A × B × C × D × E ×. There will be six gaps between
these five. Four in between and two at either end. Now if three females P, Q,R are to be arranged so that no two are together we shall
use gap method i.e., arrange them in between these 6 gaps. Hence the answer will be 6 P3 .
• Together : Suppose we have to arrange 5 persons in a row which can be done in 5 ! = 120 ways. But if two particular
persons are to be together always, then we tie these two particular persons with a string. Thus we have 5 – 2 + 1 (1
corresponding to these two together) = 3 +1 = 4 units, which can be arranged in 4! ways. Now we loosen the string and
these two particular can be arranged in 2 ! ways. Thus total arrangements = 24 × 2 = 48.
Never together = Total – Together = 120 – 48 = 72.
Example: 11 All the letters of the word ‘EAMCET’ are arranged in all possible ways. The number of such arrangement in which two
Solution: (c)
vowels are not adjacent to each other is [EAMCET 1987; DCE 2000]
Example: 12
Solution: (b) (a) 360 (b) 114 (c) 72 (d) 54
Example: 13 First we arrange 3 consonants in 3 ! ways and then at four places (two places between them and two places on two sides)
Solution: (d)
Example: 14 3 vowels can be placed in 4 P3 × 1 ways.
Solution: (a) 2!
Example: 15
Solution: (a) Hence the required ways = 3 ! × 4 P3 × 1 = 72 .
2!
Example: 16
The number of words which can be made out of the letters of the word ‘MOBILE’ when consonants always occupy odd
places is [Rajasthan PET 1999]
(a) 20 (b) 36 (c) 30 (d) 720
The word ‘MOBILE’ has three even places and three odd places. It has 3 consonants and 3 vowels. In three odd places we
have to fix up 3 consonants which can be done in 3 P3 ways. Now remaining three places we have to fix up remaining
three places which can be done in 3 P3 ways.
The total number of ways = 3 P3 ×3 P3 = 36 .
The number of 4 digit number that can be formed from the digits 0, 1, 2, 3, 4, 5, 6, 7 so that each number contain digit 1
is [AMU 2001]
(a) 1225 (b) 1252 (c) 1522 (d) 480
After fixing 1 at one position out of 4 places, 3 places can be filled by 7 P3 ways. But some numbers whose fourth digit is
zero, so such type of ways = 6 P2
∴ Total ways = 7 P3 −6 P2 = 480 .
m men and n women are to be seated in a row, so that no two women sit together. If m > n , then the number of ways in
which they can be seated is [IIT 1983]
(a) m!(m + 1)! (b) m!(m − 1)! (c) (m − 1)!(m + 1)! (d) None of these
(m − n + 1)! (m − n + 1)! (m − n + 1)!
First arrange m men, in a row in m ! ways. Since n < m and no two women can sit together, in any one of the m !
arrangement , there are (m + 1) places in which n women can be arranged in m+1 Pn ways.
∴ By the fundamental theorem, the required number of arrangement = m ! m+1 Pn = m!(m + 1)! .
(m − n + 1)!
If the letters of the word ‘KRISNA’ are arranged in all possible ways and these words are written out as in a dictionary, then
the rank of the word ‘KRISNA’ is
(a) 324 (b) 341 (c) 359 (d) None of these
Words starting from A are 5 ! = 120; Words starting from I are 5 ! = 120
Words starting from KA are 4 ! = 24; Words starting from KI are 4 ! = 24
Words starting from KN are 4 ! = 24; Words starting from KRA are 3 ! = 6
Words starting from KRIA are 2 ! = 2; Words starting from KRIN are 2 ! = 2
Words starting from KRIS are 1 ! = 1 Words starting from KRISNA are 1 ! = 1
Hence rank of the word KRISNA is 324
We are to form different words with the letters of the word ‘INTEGER’. Let m1 be the number of words in which I and N
are never together, and m2 be the number of words which begin with I and end with R. Then m1 / m2 is equal to
[AMU 2000]
(a) 30 (b) 60 (c) 90 (d) 180
Solution: (a) We have 5 letters other than ‘I’ and ‘N’ of which two are identical (E's). We can arrange these letters in a line in 5! ways.
Example: 17 2!
In any such arrangement ‘I’ and ‘N’ can be placed in 6 available gaps in 6 P2 ways, so required number = 5! 6 P2 = m1 .
2!
Now, if word start with I and end with R then the remaining letters are 5. So, total number of ways = 5! = m2 .
2!
∴ m1 = 5! . 6! . 2! = 30 .
m2 2! 4! 5!
An n digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using
only the three digits 2, 5 and 7. The smallest value of n for which this is possible is [IIT 1998]
(a) 6 (b) 7 (c) 8 (d) 9
Solution: (b) Since at any place, any of the digits 2, 5 and 7 can be used total number of such positive n-digit numbers are 3n . Since
we have to form 900 distinct numbers, hence 3n ≥ 900 ⇒ n = 7 .
Example: 18 The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy
odd places, is [Rajasthan PET 1991, 92, 98]
(a) 24 (b) 18 (c) 12 (d) 30
Solution: (b) The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places, in 4! = 6 ways and 3 even digits 2, 4, 2 can be arranged
2!2!
in the three even places 3! = 3 ways. Hence the required number of ways = 6 × 3 = 18.
2!
Circular Permutations.
So far we have been considering the arrangements of objects in a line. Such permutations are known as linear
permutations.
Instead of arranging the objects in a line, if we arrange them in the form of a circle, we call them, circular
permutations.
In circular permutations, what really matters is the position of an object relative to the others.
Thus, in circular permutations, we fix the position of the one of the objects and then arrange the other objects
in all possible ways.
There are two types of circular permutations :
(i) The circular permutations in which clockwise and the anticlockwise arrangements give rise to different
permutations, e.g. Seating arrangements of persons round a table.
(ii) The circular permutations in which clockwise and the anticlockwise arrangements give rise to same
permutations, e.g. arranging some beads to form a necklace.
Look at the circular permutations, given below :
AA
DB BD
C C
Suppose A, B, C, D are the four beads forming a necklace. They have been arranged in clockwise and
anticlockwise directions in the first and second arrangements respectively.
Now, if the necklace in the first arrangement be given a turn, from clockwise to anticlockwise, we obtain the
second arrangement. Thus, there is no difference between the above two arrangements.
(1) Difference between clockwise and anticlockwise arrangement : If anticlockwise and clockwise
order of arrangement are not distinct e.g., arrangement of beads in a necklace, arrangement of flowers in garland
etc. then the number of circular permutations of n distinct items is (n − 1)!
2
(2) Theorem on circular permutations
Theorem 1 : The number of circular permutations of n different objects is (n − 1)!
Theorem 2 : The number of ways in which n persons can be seated round a table is (n − 1)!
Theorem 3 : The number of ways in which n different beads can be arranged to form a necklace, is
1 (n − 1)! .
2
Note : When the positions are numbered, circular arrangement is treated as a linear arrangement.
In a linear arrangement, it does not make difference whether the positions are numbered or not.
Example: 19 In how many ways a garland can be made from exactly 10 flowers [MP PET 1984]
Solution: (d) (a) 10 ! (b) 9 ! (c) 2 (9!) (d) 9!
Example: 20 2
Solution: (b)
Example: 21 A garland can be made from 10 flowers in 1 (9!) ways [ n flower's garland can be made in 1 (n − 1)! ways]
Solution: (a) 2 2
Example: 22 In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit together [IIT 1975; MP PET 1987]
Solution: (a)
(a) 5! × 5! (b) 4! × 5 ! (c) 5!× 5! (d) None of these
2
Since total number of ways in which boys can occupy any place is (5 − 1)!= 4! and the 5 girls can be sit accordingly in 5!
ways. Hence required number of ways are 4 ! × 5 !.
The number of ways in which 5 beads of different colours form a necklace is [Rajasthan PET 2002]
(a) 12 (b) 24 (c) 120 (d) 60
The number of ways in which 5 beads of different colours can be arranged in a circle to form a necklace are
= (5 − 1)!= 4! .
But the clockwise and anticlockwise arrangement are not different (because when the necklace is turned over one gives rise
to another). Hence the total number of ways of arranging the beads = 1 (4! ) = 12 .
2
The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that
the two female are not seated together is [Roorkee 1999]
(a) 480 (b) 600 (c) 720 (d) 840
Fix up a male and the remaining 4 male can be seated in 4! ways. Now no two female are to sit together and as such the 2
female are to be arranged in five empty seats between two consecutive male and number of arrangement will be 5 P2 .
Hence by fundamental theorem the total number of ways is = 4! × 5 P2 = 24 × 20 = 480 ways.
Combinations
Definition.
Each of the different groups or selections which can be formed by taking some or all of a number of objects,
irrespective of their arrangements, is called a combination.
Suppose we want to select two out of three persons A, B and C.
We may choose AB or BC or AC.
Clearly, AB and BA represent the same selection or group but they give rise to different arrangements.
Clearly, in a group or selection, the order in which the objects are arranged is immaterial.
Notation: The number of all combinations of n things, taken r at a time is denoted by C (n, r) or n Cr or nr .
(1) Difference between a permutation and combination : (i) In a combination only selection is made
whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered.
(ii) In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering
is essential. For example A, B and B, A are same as combination but different as permutations.
(iii) Practically to find the permutation of n different items, taken r at a time, we first select r items from n items
and then arrange them. So usually the number of permutations exceeds the number of combinations.
(iv) Each combination corresponds to many permutations. For example, the six permutations ABC, ACB,
BCA, BAC, CBA and CAB correspond to the same combination ABC.
Note : Generally we use the word ‘arrangements’ for permutations and word “selection” for
combinations.
Number of Combinations without Repetition.
The number of combinations (selections or groups) that can be formed from n different objects taken
r(0 ≤ r ≤ n) at a time is nCr = n!
r !(n − r)!
Let the total number of selections (or groups) = x. Each group contains r objects, which can be arranged in r ! ways.
Hence the number of arrangements of r objects = x ×(r!) . But the number of arrangements = n Pr .
⇒ x × (r!) = n Pr ⇒x= n Pr ⇒ x = n! =nCr .
r! r !(n − r)!
Important Tips
• nCr is a natural number. n C0 =nCn = 1, nC1 = n
• n Cr =nCn−r n Cr + nCr−1 =n+1Cr
• nCx =nCy ⇔ x = y or x + y = n n. n−1Cr−1 = (n − r + 1)n Cr−1
• If n is even then the greatest value of n Cr is n Cn / 2 . If n is odd then the greatest value of nCr is n Cn+1 or n Cn−1 .
2 2
• n Cr = n .n −1 Cr −1 n Cr = n−r +1
r n Cr −1 r
• nC0 +nC1 +nC2 + ..... +nCn = 2n nC0 +nC2 +n C4 + ...... =nC1 +nC3 +nC5 + ..... = 2n−1
• 2n+1C0 +2n+1C1 +2n+1C2 + ..... +2n+1Cn = 22n nCn +n+1Cn +n+2Cn +n+3Cn + .... +2n−1Cn =2nCn+1
Note : Number of combinations of n dissimilar things taken all at a time n Cn = n! = 1 = 1, (0 ! = 1) .
n !(n − n)! 0!
Example: 23 If 15 C3r =15Cr+3 , then the value of r is [IIT 1967; Rajasthan PET 1991; MP PET 1998; Karnataka CET 1996]
Solution: (a) (a) 3 (b) 4 (c) 5 (d) 8
Example: 24
15 C3r =15Cr+3 ⇒ 15 C15−3r =15Cr+3 ⇒ 15 − 3r = r + 3 ⇒ r = 3 .
n Cr = [MP PET 1984]
n Cr−1
(a) n−r (b) n+r −1 (c) n−r +1 (d) n−r −1
r r r r
Solution: (c) n Cr = n! ⇒ n! × (r − 1)!(n − r + 1)! = (n − r+ 1)(r − 1)! (n − r)! = (n − r + 1) .
n Cr −1 r!(n − r)! r!(n − n! r(r − 1)!(n − r)! r
r)!
n!
(r − 1)!(n − r + 1)!
Example: 25 If n+1 C3 = 2nC2 , then n = [MP PET 2000]
Solution: (c)
(a) 3 (b) 4 (c) 5 (d) 6
n+1 C3 = 2.n C2
⇒ (n + 1)! = 2. n! 2)! ⇒ n+1 = 2 ⇒ n+1= 6⇒n= 5.
3!(n − 2)! 2!(n − 3.2! 2!
Example: 26 If n Cr−1 = 36, nCr = 84 and n Cr+1 = 126 then the value of r is [IIT 1979; Pb. CET 1993; DCE 1999; MP PET 2001]
(a) 1 (b) 2 (c) 3 (d) None of these
Solution: (c) Here n Cr−1 = 36 and n Cr = 84
n Cr 84 n Cr+1 126
3n − 10r = −3 and 4n − 10r = 6 ; on solving we get n = 9 and r = 3 .
Example: 27 In a conference of 8 persons, if each person shake hand with the other one only, then the total number of shake hands
shall be [MP PET 1984]
(a) 64 (b) 56 (c) 49 (d) 28
Solution: (d) Total number of shake hands when each person shake hands with the other once only = 8 C2 = 28 ways.
Example: 28 How many words of 4 consonants and 3 vowels can be formed from 6 consonants and 5 vowels. [Rajasthan PET 1985]
(a) 75000 (b) 756000 (c) 75600 (d) None of these
Solution: (b) Required number of words = 6 C4 ×5 C3 × 7! = 756000
[Selection can be made in 6 C4 × 5 C3 while the 7 letters can be arranged in 7!]
Example: 29 To fill 12 vacancies there are 25 candidates of which five are from scheduled caste. If 3 of the vacancies are reserved for
scheduled caste candidates while the rest are open to all, then the number of ways in which the selection can be made
[Rajasthan PET 1981]
(a) 5 C3 ×22 C9 (b) 22 C9 −5 C3 (c) 22 C3 + 5 C3 (d) None of these
Solution: (a) The selection can be made in 5 C3 ×22 C9 [since 3 vacancies filled from 5 candidates in 5 C3 ways and now remaining
candidates are 22 and remaining seats are 9, then remaining vacancies filled by 22 C9 ways. Hence total number of ways
5 C3 ×22 C9 .
Number of Combinations with Repetition and All Possible Selections.
(1) The number of combinations of n distinct objects taken r at a time when any object may be repeated any
number of times.
= coefficient of x r in (1 + x + x 2 + ....... + x r )n = coefficient of x r in (1 − x)−n =n+r−1Cr
(2) The total number of ways in which it is possible to form groups by taking some or all of n things at a time
is 2n − 1.
(3) The total number of ways in which it is possible to make groups by taking some or all out of
n = (n1 + n2 + ....) things, when n1 are alike of one kind, n2 are alike of second kind, and so on is
{(n1 + 1)(n2 + 1)......} − 1.
(4) The number of selections of r objects out of n identical objects is 1.
(5) Total number of selections of zero or more objects from n identical objects is n + 1 .
(6) The number of selections taking at least one out of a1 + a2 + a3 + ...... + an + k objects, where a1 are alike
(of one kind), a2 are alike (of second kind) and so on...... an are alike (of nth kind) and k are distinct =
[(a1 + 1)(a2 + 1)(a3 + 1).......(an + 1)] 2k − 1 .
Example: 30 There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall
Solution: (b) can be illuminated is [Roorkee1990]
Example: 31
Solution: (a) (a) 102 (b) 1023 (c) 210 (d) 10 !
Example: 32 Number of ways are = 210 − 1 = 1023
Solution: (d) [– 1 corresponds to none of the lamps is being switched on.]
Example: 33
Solution: (b) 10 different letters of English alphabet are given. Out of these letters, words of 5 letters are formed. How many words are
formed when atleast one letter is repeated [UPSEAT 1999]
(a) 99748 (b) 98748 (c) 96747 (d) 97147
Number of words of 5 letters in which letters have been repeated any times = 105
But number of words on taking 5 different letters out of 10 = 10 C5 = 252
∴ Required number of words = 105 − 252 = 99748.
A man has 10 friends. In how many ways he can invite one or more of them to a party [AMU 2002]
(a) 10 ! (b) 210 (c) 10 ! − 1 (d) 210 − 1
Required number of friend = 210 − 1 (Since the case that no friend be invited i.e., 10 C0 is excluded)
Numbers greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits
is allowed), are [AIEEE 2002; IIT 1976]
(a) 350 (b) 375 (c) 450 (d) 576
Numbers greater than 1000 and less than or equal to 4000 will be of 4 digits and will have either 1 (except 1000) or 2 or 3
in the first place with 0 in each of remaining places.
After fixing 1st place, the second place can be filled by any of the 5 numbers. Similarly third place can be filled up in 5 ways
and 4th place can be filled up in 5 ways. Thus there will be 5 × 5 × 5 = 125 ways in which 1 will be in first place but this
include 1000 also hence there will be 124 numbers having 1 in the first place. Similarly 125 for each 2 or 3. One number
will be in which 4 in the first place and i.e., 4000. Hence the required numbers are 124 + 125 + 125 +1 = 375 ways.
Conditional Combinations.
(1) The number of ways in which r objects can be selected from n different objects if k particular objects are
(i) Always included = n−k Cr−k (ii) Never included = n−k Cr
(2) The number of combinations of n objects, of which p are identical, taken r at a time is
= n−p Cr + n−pCr−1 + n−pCr−2 + ....... + n−pC0 if r ≤ p and
= n−p Cr + n−p Cr−1 + n−p Cr−2 + ....... + n−p Cr−p if r > p
Example: 34 In the 13 cricket players 4 are bowlers, then how many ways can form a cricket team of 11 players in which at least
Solution: (c)
2 bowlers included [Rajasthan PET 1998]
(a) 55 (b) 72 (c) 78 (d) None of these
The number of ways can be given as follows:
2 bowlers and 9 other players = 4 C2 ×9C9 ; 3 bowlers and 8 other players = 4 C3 ×9C8
4 bowlers and 7 other players = 4 C4 ×9C7
Hence required number of ways = 6 × 1 + 4 × 9 + 1 × 36 = 78.
Example: 35 In how many ways a team of 10 players out of 22 players can be made if 6 particular players are always to be included
Solution: (c)
Example : 36 and 4 particular players are always excluded [Rajasthan PET 2002]
Solution: (c)
(a) 22 C10 (b) 18 C3 (c) 12 C4 (d) 18 C4
6 particular players are always to be included and 4 are always excluded, so total number of selection, now 4 players out
of 12.
Hence number of ways = 12C4 .
In how many ways can 6 persons to be selected from 4 officers and 8 constables, if at least one officer is to be included
(a) 224 (b) 672 (c) 896 [Roorkee 1985; MP PET 2001]
(d) None of these
Required number of ways = 4 C1 × 8C5 +4C2 × 8C4 +4C3 ×8C3 + 4C4 × 8C2 = 4 × 56 + 6 × 70 + 4 × 56 + 1 × 28 = 896.
Division into Groups.
Case I : (1) The number of ways in which n different things can be arranged into r different groups is n+r−1 Pn
or n ! n−1 Cr−1 according as blank group are or are not admissible.
(2) The number of ways in which n different things can be distributed into r different group is
r n −r C1(r − 1)n +r C2(r − 2)n − ......... + (−1)n−1 nCr−1 or Coefficient of x n is n ! (e x − 1)r
Here blank groups are not allowed.
(3) Number of ways in which m × n different objects can be distributed equally among n persons (or numbered
groups) = (number of ways of dividing into groups) × (number of groups) ! = (mn)! n ! = (mn)! .
(m!)n n! (m!)n
Case II : (1) The number of ways in which (m + n) different things can be divided into two groups which
contain m and n things respectively is, m+n Cm .n Cn = (m + n)! , m ≠ n.
m ! n!
Corollary: If m = n , then the groups are equal size. Division of these groups can be given by two types.
Type I : If order of group is not important : The number of ways in which 2n different things can be
divided equally into two groups is (2n)!
2!(n!)2
Type II : If order of group is important : The number of ways in which 2n different things can be divided
equally into two distinct groups is (2n)! × 2! = 2n!
2!(n!)2 (n!)2
(2) The number of ways in which (m + n + p) different things can be divided into three groups which contain
m, n and p things respectively is m+n+p Cm .n+pCn .p Cp = (m + n + p)! ,m ≠ n≠ p
m!n! p!
Corollary: If m = n = p , then the groups are equal size. Division of these groups can be given by two types.
Type I : If order of group is not important : The number of ways in which 3p different things can be
divided equally into three groups is (3 p)!
3!(p!)3
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