DUM10122 WORKBOOK

DUM 10122: Engineering Mathematics 1

7. Given that loga 2 = 0.631 and loga 5 = 1.465, calculate the following :

a) loga 20 b) loga 50

c) loga 8 d) loga  2 
5

e) loga 3 2 f) loga 10

8. Given that loga 2 = 0.501 and loga 3 = 0.794, find the value of :

a) loga 9 b) loga 24

20

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Convert a number in index form to the logarithmic form and vice
versa.

4.3 CONVERSION BETWEEN INDICIAL FORM AND LOGARITHMIC FORM

We can solve equations such as x = 25 and x2 = 5 directly. But in cases such as 5x = 2, we
cannot solve the equation directly. Therefore, we convert it to logarithmic form. By definition,

ax = y  loga y = x,

where a > 0 and a  1. Hence, an equation in the index form can be converted to the
logarithmic form and vice versa.

Write 125 = 53 as a logarithmic form.
Solution :

25 = 52 , therefore, log5 25 = 2.

Express log2 32 = 5 in index form.
Solution :

log2 32 = 5, Therefore 25 = 32.

21

DUM 10122: Engineering Mathematics 1

TUTORIAL 4.3

1. Express the following in logarithmic form :

a) 72 = 49 1

b) 16 2 = 4

c) 10-3 = 0.001 d)  1 −2 = 4
e) 52 = 25 2

f) 512 = 83

1 h) 2x = y

g) 22 = 2

2. Express the following in index form :

a) log4 2 = 1 b) log9 27 = 3
2 2

c) log10 1 =- 2 d) log3 3=1
100 2

e) 4 = log2 16 f) x = log3 y

22

DUM 10122: Engineering Mathematics 1

After completing the unit, students should be able to:

1. Solve indicial by comparison of the base or index

UNIT LEARNING • using laws of indices
OUTCOMES
• by changing to logarithmic form

4.4 SOLUTIONS OF IN INDICIAL EQUATIONS

The are two ways of solving equations involving indices.

Case 1 Unknown on the indexs, let the base have the same value.
Case 2
ax =ay

By comparison index, x = y

Unknown on the base, let the indexs have the same value.

ax =bx

By comparison base, a = b

Case 3 Change indicial form to logarithmic form.

Solve the following indicial equations.

a) 5x = 54

Solution :
x = 4

b) x3 = 43
Solution :

x = 4

c) 2x4 = 32

Solution :
x4 = 16
x4 = 24
x = 2

23

DUM 10122: Engineering Mathematics 1

d) 2x −1 = 1
64

Solution :
2x – 1 = 2-6
x – 1 = -6
x = -5

e) 5x = 4

Solution :
log 5x = log 4

x log 5 = log 4

x = log 4
log 5

= 0.8614

a) 5-6 = (5x)-6 Solve the following indicial equations :
b) 32x-1 = 243

24

TUTORIAL 4.4 DUM 10122: Engineering Mathematics 1
Solve the following indicial equations : 2)  1 4n =  1 −6

1) 3n = 37 5 5

3) e2 = e 1x 4) (-2)3 = (-2)6n
2

5) n5 = 35 6) 5-6 = (5x)-6

7) 2x x 3x = 36

8) 8x3 = 27

9) 5x + 2 = 25 10) 72x – 1 – 49 = 0

25

11) 6 2x = 7776 DUM 10122: Engineering Mathematics 1
6x 12) 33x = 81 x 3x

13) 8x = 128 14) 252n = 625

15) 43x – 1 = 1 16) 243(33x) = 1
17) 82x = 1 18) 25x = 1

32 5

26

19) 493x – 1 = 1 DUM 10122: Engineering Mathematics 1
343
20) 1 =81
3x

1 1

21) x 2 = 6 22) x 3 = 4

1 1

23) 2x 2 =18 24) (x + 4)2 = 4

25) (3m)4 = 16 26) (6r)5 = 32

27

27) (2x + 1)3 = 64 DUM 10122: Engineering Mathematics 1
28) (5n – 1)2 = 81

2 30) 53x  25x +1 = 1
125
29) 7x 3 − 5 = 23

1 32) 1 x3 =128
4
31) 2x 2 = 4x

28

33) 2x = 10 DUM 10122: Engineering Mathematics 1
34) 52x = 15

35) 3x + 1 = 12 36) 32x -1 = 13

37) ex = 2 38) e2x + 1 = 5

29

DUM 10122: Engineering Mathematics 1

UNIT LEARNING After completing the unit, students should be able to:
OUTCOMES
1. Solve logarithmic equations:
• by comparison of the base
• using law of logarithms

• by changing to index form.

4.5 SOLUTIONS OF LOGARITHMIC EQUATIONS

Solve the following logarithmic equations.

a) log2 x = log2 14

Solution :
log2 x = log2 14
Since the bases are the same, the numbers are equal.
 x = 14

b) log8 2x + log8 2 = 3log8 2

Solution :
log8 (2x)(2) = log8 23
log8 4x = log88
x=2

c) log3 (2x – 21) – log3 (x – 7) = 2

First Method

log3 (2x – 21) – log3 (x – 7) = 2
log3 (2x – 21) – log3 (x – 7) = 2 log3 3
log3 (2x – 21) – log3 (x – 7) = log3 32

log3 2x − 21 = log3 9
x−7

Since the bases are the same, the numbers are equal.

 2x − 21 = 9
x−7

2x – 21 = 9(x – 7)
2x – 21 = 9x – 63
9x – 2x = -21 + 63

7x = 42
x=6

30

DUM 10122: Engineering Mathematics 1

Second Method

log3 (2x – 21) – log3 (x – 7) = 2

log3 2x − 21 =3
x−7

Change to index form

2x − 21 = 32
x−7

2x – 21 = 9(x – 7)
2x – 21 = 9x – 63

9x – 2x = -21 + 63
7x = 42
x=6

TUTORIAL 4.5 2) log (2x – 1) = log 4

Solve the following logarithmic equations :
1) log4 2x = log4 84

3) log2 (x – 3) = log2 2x 4) log5  1  = log5 10
 2x 

31

5) log2 (x2 – 3) = log2 2x DUM 10122: Engineering Mathematics 1
7) log2 18 = log2 2x2 6) ln x2 = ln 64

8) log 2x2 = log (4 – 2x)

9) ln (2x – 4) = ln 4 – ln 2 10) log x = log 4 + log 2

11) log5 x – log5 (x – 5) = log5 3 12) 2 log x = log 32x + log 2

32

13) log5 x = 2 DUM 10122: Engineering Mathematics 1
15) logx 27 = 3 14) log x = 2
17) x = log3 9
16) logx 2 = 1
2

18) x = log4 8

19) ln x = 3 20) 2 ln x = 8
21) logx 100 = 2 22) log 1000 = x

33

23) log (4x – 3) + log 5 = 1 DUM 10122: Engineering Mathematics 1
24) ln x + ln 2 = 2

25) log2 (x – 1) – log2 (x +1) = 2 26) log3 x – log3 (2x – 1) – 3 = 0

27) log4 3 = x 28) log 3 8 = x

34

29) 2 log 3 = x DUM 10122: Engineering Mathematics 1
30) logx 3 = 1.5

QUICK CHECK !

35

DUM 10122: Engineering Mathematics 1

ANSWERS TUTORIAL 4.4

TUTORIAL 4.1 1) n = 7, 2) n = − 3 , 3) x = 4, 4) n = 1 ,
2 2

1. a) 36 + 3n b) 5-3n c) 125 d) 9 5) n = 3, 6)x = 1, 7)x = 2, 8) x = 3 9) x = 0,
e) 5 6 – 2n f) 3n + 7 g) 2h-2 2
h) 2m -3 i) 3n-1 j) 26n+9 k) 25n+1
l) k-3-m m) 38x – 14 n) 128a9b5 10)x = 3 , 11)x = 5, 12) x = 2, 13) x = 7 ,
2 3

TUTORIAL 4.2 14) n = 1, 15) x = 1 , 16) x = − 5 , 17) x = − 5 ,
3 3 6
1 –5
1 a) 8loga x – 4loga y b) 2 loga y loga x, 18) x = − 1 , 19) x = − 1 ,
2 2 6

c) 4 loga x – 2 loga y, d)– 1 loga x, 20) x = -4, 21) x = 36, 2) x = 64, 23) x = 81,
3 3 4
e) 4loga x + loga y +5 loga z, 24) x = 12, 25) m = 2 , 26) r = 1 , 27) x = 3 ,
3 3 2
f) loga x + loga y – loga z, g) loga x – 2loga y –
logaz, 1
h) loga x + 2loga y – loga z 28) n = 2, 29) x = 8, 30) x = -1, 31) x = 4 ,

32) x = 8 , 33) 3.3219, 34) 0.8413, 35) 1.2619,

2. a) loge  25  ,b) logp 12 c) log2 64 or 6 36) 1.6674, 37) 0.6931, 38) 0.3047
 
t

,d)log3  x 32  , e) log  x 2 y  ,f) log  ab 3  TUTORIAL 4.5
 y 15  c2
  z

g) ln xy4 , h) ln xy 1) 42, 2) 5 , 3) -3, 4) 1 ,5) 3 or -1, 6) 8
2 20

3. a) 1, b) -2, c) 4, d) -3, e) 2, f) 2 7)  3 ,8) -2 or 1, 9) 3, 10) 8, 11) 15 , 12) 64, ,
2
4. a) 2.2619, b) 2.93 c) 0.6910, d) 1.3571
13) 25, 14) 100, 15) 3, 16) 4, 17) 2, 18) 3 ,
5. a) 1.97, b) 2.54, c) -0.75, d) 0.555,
2

6. a) 2r + 1, b) r –1 , c) 1 , d) -3r, 19) 20.09, 20) 54.598, 21) 10, 22) 3, 23) 5 ,
r
4

7. a) 2.727, b) 3.561, c) 1.893, d) -0.834, 24) 3.6945, 25) − 5 , 26) 27 , 27) 0.7925,
e) 0.2103, f) 1.048 53
8. a) 1.588, b) 2.297 3
28) 1.8928, 29) 0.9542, 30) 2.08

36

Unit 5 DUM 10012 Engineering Mathematics 1
COMPLEX NUMBERS
Unit 5: Complex Numbers

INTRODUCTION OVERVIEW

Complex number is a combination of real numbers and imaginary numbers. It’s occurred to

solve a square root of negative numbers. Thus, we imagine that does exist. Hence,

Complex numbers often seem strange when first encountered but it is worth preserving with
them because they provide a powerful mathematical tool for solving several technical
problems. One of the main applications is to the analysis of alternating current (a.c) circuits.
Engineers are very interested in these because the main supply itself a.c, and electricity
generation and transportation are dominated by a.c voltages and currents.

Complex does not mean complicated. It means the two types of numbers, real and
imaginary, together from a complex, just like a building complex (buildings joined together).

After completing this unit, students should be able to :

1. Write the square root of negative numbers in terms of i

2. Simplify powers of i

3. State complex number in rectangular form. UNIT LEARNING
4. Plot complex number in Argand Diagram. OUTCOMES

5. Perform operations of complex numbers.

6. Perform division of complex numbers using the conjugate.

7. Calculate the modulus and argument of a complexnumber.

8. Convert complex number in rectangular form to polar form and vice versa.

9. Write complex number in polar form and trigonometric form.

10. Use De Moivre’s Theorem to find powers of a complex number.

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

CONTENT CONTENT
5.1 Imaginary Numbers
5.2 Complex Numbers in Rectangular Form and Argand Diagram
5.3 Complex Numbers in Polar Form
5.4 De Moivre’s Theorem

3

5.1 IMAGINARY NUMBERS DUM 10012 Engineering Mathematics 1

IMAGINARY NUMBER Unit 5: Complex Numbers

➢ An imaginary number is a number that can ➢ Imaginary numbers also look like − 3 or
be written as a real number multiplied by − 5 . We can use the rules for simplifying
the imaginary unit i, which is defined by its
property i2 = -1. square roots to rewrite these kinds of
imaginary numbers for example:
➢ We define the imaginary numbers as the
square root of -1, and represent it as i. a) − 3 = 3(−1) = 3  −1 = 3i

IMAGINARY NUMBER b) − 5 = 5(−1) = 5  −1 = 5i

i = −1 ➢ These can generate formula for simplifying
which square roots as imaginary numbers
i2 = −1
GENERAL FORMULA

− a = a(−1) = a  −1 = ai

Example : Write as imaginary numbers for the following square roots.

Convert the following to imaginary number. Try Convert the following to imaginary number.
This
a) − 4 a) −9
b) − −16
b) − − 25
Solution:

a) − 4 = 4  −1 = 2i

b) − − 16 =- 16  − 1 =-4i

4

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

SIMPLIFYING IMAGINARY NUMBER

POWER OF i

i = −1 . .
i2 = i i = −1 −1 = −1

i3 = i i i = − 1 − 1 − 1 = −i

i4 = i i i i = − 1 −1 −1 −1 = 1

In fact, for every factor of i4, can be replaced by the value of 1.
i4 = 1

How to simplify i to any given power:
Since i4 is the number 1 we can apply the following formula to reduce i to any power

ik = ir
where r = the remainder of k ÷ 4

Example : Simplifying imaginary number.

Simplify the following Simplify the following

a) i9 Try
b) i22
Thi s a) i27
b) i104

Solution:

a) 2
i9 → 4 9 → remainder = 1

−8

1

i9 = i1 = √−1

b) 5
i22 → 4 22 → remainder = 2

− 20

2

i22 = i2

5

Example : Simplifying imaginary number. DUM 10012 Engineering Mathematics 1

Evaluate the following Try Unit 5: Complex Numbers

a) i103 This Evaluate the following
ia) 46
b) i28 ib) 81
Solution:

a) 25
i103 → 4 103 → remainder = 3

− 100

3

i103 = i3 = −i

b) 7
i28 → 4 28 → no remainder

− 28

0

i28 = i4 = 1

Example : Differentiate between imaginary numbers and real numbers

Determine which of the following is the imaginary Try
numbers. This

a. 20i a. 2

b. − 81 b. √−6
c. √60
c. - 2 d. -6
d. 40 e. 3
e. – 8 f. √1

f. − 81 2

g. -13 i g. √− 1
h. -5i
4

h. 56
i. -13 i
j. 5i

6

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

TUTORIAL 5.1

1. Convert the following square root to complex number

a) − 81 b) − 49

c) − − 36 d) − − 64

e) − 9 f) − −100

g) − − 144 h) − 400

i) − 10 j) − 14

2. Simplify each of the following. b) i32
a) i30

c) i77 d) i51

e) i308 f) i241

7

3. Evaluate each of the following. DUM 10012 Engineering Mathematics 1
a) i9
Unit 5: Complex Numbers

b) i12

c) i7 d) i96

e) 14 f) i201

4. Determine which of the following an imaginary number.

a. -3
b. − √−2
c. 30
d. -12i
e. − √15
f. √−4
g. 15

8

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

5.2 COMPLEX NUMBERS IN RECTANGULAR FORM AND ARGAND DIAGRAM

5.2.1 Complex Numbers In Rectangular Form ➢ Complex numbers in rectangular form
is any number, real or imaginary that
STANDARD FORM OF COMPLEX can be written in the form
NUMBER
a + bi ➢ Where a and b are real numbers and i is the
imaginary number. This form, a + bi, is
called the standard form of a complex
number.

Example: Identify real part and imaginary part in rectangular form.

Write the following number in rectangular form Write the following number in rectangular form

4+ −4 Try a) 3 − − 81
This b) − − 25 + 2
Solution:
c) − 16 −3
4 + − 4 = 4 + 4  −1

= 4+ 4  −1
= 4 + 2i

8

5.2.2 Argand Diagram DUM 10012 Engineering Mathematics 1
bi (imaginary axis)
Unit 5: Complex Numbers
(-ve , +ve) (+ve , +ve)
(-ve , -ve) a (real axis) ➢ Given a complex number : Z = a +bi
➢ We can obtain a useful graphical
(+ve , -ve)
interpretation of it by plotting the real part
on the horizontal axis and the imaginary
part on the vertical axis.
➢ We call the horizontal axis as the real axis
and vertical axis as an imaginary axis.

9

DUM 10012 Engineering Mathematics 1

Example : To plot complex numbers in the Argand Diagram Unit 5: Complex Numbers

Plot the complex number below using the Argand Plot the complex number below using the
Diagram.
Try

This Argand Diagram. Plot in one arganddiagram.

a) Z1 = 2 + 4i a) Z1 = −5 + i

b) Z2 = 4 − 3i b) Z2 = −7 + 6i

c) Z3 = −3 −3i

Solution:

bi

4 -- Z1 (2,4)

| || a

-3 2 4

-3 – Z2 (4,-3)
Z3 (-3,-3)

5.2.3 Operations Of Complex Numbers

Addition, Substraction And Multiplication Of Complex Numbers

➢ Given Z1 = (a +bi)and Z2 = (c +di)

ADDITION

Z1 + Z2 = (a +bi)+(c + di)= (a + c)+(b +d) i

“FOIL” SUBSTRACTION

Z1 −Z2 = (a +bi)−(c +di) = (a −c)+(b −d) i

10

First, Outers, Inners, Lasts DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

MUTIPLICATION

Z1Z2 = (a +bi)(c +di) = (ac −bd)+(ad +bd) i

Example : To add and subtract complex numbers

If Z1 = 4 + 8i and Z2 = 5 − 3i , calculate: Try If Z1 = 7 + i and Z2 = 3 − 5i , calculate:
This a) Z1 +Z2
a) Z1 +Z2
b) Z1 −Z2 b) Z2 − Z1

Solution: Solution:

a)

Z1 + Z2 = (4 + 8i)+ (5 − 3i)
= (4 + 5)+ (8−3)i

= 9 + 5i

b)

Z1 − Z2 = (4 + 8i)− (5−3i)
=(4−5)+ (8+3)i

= −1+ 11i

11

Example : To multiply complex numbers DUM 10012 Engineering Mathematics 1

If Z1 = 4 − 3i and Z2 = 3 + 6i , calculate: Try Unit 5: Complex Numbers
a) 4Z1 This
If Z1 = 6 + 2i and Z2 = −1+ i , calculate:
1 a) − 0.5 Z
b) 3Z1
1

2
b) 5 Z2

c) Z1Z2 c) Z1Z2

Solution: Solution:
a)
4Z1 = 4(4−3i)
b)
= 16 −12i
c)
1 Z = 1 (3 + 6i)

32 3
= 1+ 2i

Z1Z2 = (4 − 3i)(3 + 6i)

= 12 + 24i − 9i −18i2

= 12 + 15i −18(−1)

= 12 + 15i + 18
= 30 + 15i

Division Of Complex Number Using The Conjugate

Conjugate ➢ A conjugate is where we change the
sign of imaginary part of complex
CONJUGATE OF COMPLEX NUMBER number

a + bi = a −bi ➢ A conjugate is often written with a bar
over it given

12

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

Example : To obtain the conjugate of complex numbers
Find the conjugate of the following complex
Find the conjugate of the following complex numbers. Try

a) Z = 2 + 3i This numbers.

b) Z = i − 5 a) Z = 14 − 9i

c) Z = −2i+ 6 b) Z = −i + 8

Solution: c) Z = 7i −1

a) Z = 2 − 3i Solution:

b) Z = −i− 5

c) Z = 2i+ 6

Division Of Complex Numbers

DIVISION COMPLEX NUMBER

Given Z1 = (a + bi)
Z2 (c + di)
Z1 = (a +bi) and Z2 = (c +di)
= ((ac++bdii))((cc−−ddii))

= (a + bi)(c − di)

c2 + d2

= (ac + bd) + (bc − ad) i
+
c2 d2 c2 + d2 

Example : To divide complex numbers

If Z1 = 2 + 6i and Z2 = 1− 4i , calculate: Try If Z1 = 3 −i and Z2 = 9 −12i calculate:
Z1 This Z2

a) 2 a) 3
Z1 Z2

b) Z 2 b) Z1

Solution: Solution:

13

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

a)

Z1 = 2 + 6i
2 2

= 1+3i

b)

Z1 = 2+ 6i
Z2 1− 4i

= 2+ 6i  1+ 4i
1− 4i 1+ 4i

= 2 + 8i + 6i + 24i2
12 +42

= 2 + 14i + 24(−1)

17

= 2 − 24 + 14i
17

= − 22 + 14i
17

= − 22 + 14 i
17 17

14

TUTORIAL 5.2 Real Part DUM 10012 Engineering Mathematics 1
1. Complete the table below:
Unit 5: Complex Numbers
Complex Number
Imaginary Part
2 + 3i
– 7i
7−i i

− 5 + 3i – 9i

4
1
–8

2. Write in rectangular form the following numbers.

a) 2 + − 4 b) −4 + −16

c) 3 + − 25 d) 1+ − 49

e) 1− − 9 f) 5 − − 81

3. Plot the complex number below using the Argand Diagram:

a) z1 = −5 − 3i b) z2 = 4 −i

c) z3 = 6 + i d) z4 = 3 −7i

15

e) z5 = 2 −5i DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

f) z6 = 2 −8i

4. Find the conjugate of the following complex numbers.

a) 3 + 2i b) 5i −18

c) 1−10i d) −i + 6
b) z1 − z2
5. If z1 = 3 + 4i and z2 = −2 + i, find
a) z1 + z2

c) z1z2 z1
d) z2

16

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

6. If z1 = 2 + 7i , z2 = 1− 4i and z3 = −3 + 3i , find

a) z1 + z2 b) z3 − z1

c) z1 + z2 − z3 d) z1z2

17

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

e) z3 f) z2
z1 z3

7. If z1 =1− 3i, z2 = 7i − 8 and z3 = 2 + 4i , find

a) z1 + z2 b) z2 − z3

c) z1 z2 d) z2 z3

18

e) z1z3 DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

f) z 3
z1

8. In an alternating circuit, the total voltage V is given by V = V1 + V2 . If V = (12.2 + 6.8i)V

and V1 = (7.8 − 2.5i)V, calculate the voltage V2. JD 18’

9. In an alternating current circuit containing two impedances z1 and z2 in parallel, the total

impedance z is given by the formula z = z1z2 . If z1= (4+ 5i)  and z =2 (7−3i) ,
z1 + z2

find the total impedance, z

19

5.3 DUM 10012 Engineering Mathematics 1

bi Unit 5: Complex Numbers

COMPLEX NUMBERS IN POLAR FORM

The Modulus And Argument Of Complex Number

Z = a + bi ➢ From figure 4.2, by using Pythagoras’
theorem we obtain modulus, r.
|Z| = r
➢ Clearly, r is the distance of the point (a,
b b) from the origin.

Ɵ a ➢ The modulus is always a non-negative
a number and is denoted Z .

Figure 4.2 ➢ The angle is conventionally measured
from the positive x axis. The angle, is
MODULUS called the argument of Z.

r = a2 + b2 ARGUMENT

 = tan−1ab

Quadrant 2 : bi
Quadrant 1 :

Argument Argument a
Argument Argument

Quadrant 3 : Quadrant 3 :

Principal value :
Figure 4.3

20

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

Example : To determine modulus, r andargument,

Find the modulus and argument from the given Try Find the modulus and argument from
Argand diagram below the given Argand diagram below

This

Solution: Solution :

bi
bi

a

a

a) Modulus,
r = a2 + b2

= 22 + 42

= 4.47  = tan−1 4  = 63.43
Basic angle,

 2

Z1 in quadrant 1. Therefore argument,

 = 63.43 Solution:

Solution : b

bi a

a

b) Modulus,

r = a2 + b2

= (− 1)2 + 12

= 1.41  = tan−1 1 = 45

Basic angle,

1
Z2 in quadrant 2. Therefore argument,

 = 180 − 45 = 135

21

a) z = 2 + 2i DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

b) z = 5 +12i

5.3.2 Complex Number In Rectangular Form and Polar Form (Trigonometric Form)

RECTANGULAR FORM ➢ The polar form of the complex number a +bi

a + bi is r (cos +isin) where,

POLAR FORM / TRIGONOMETRIC r = a2 + b2 and  = tan−1 b 
FORM  
 a 
a+bi = r (cos+isin)
➢ Another notation for polar form is
Or
r .
a + bi=r
(This is spoken as “r at angle  ”.)

➢ The length r is called the magnitude of the

complex number, and the angle  is called the

argument.

22

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

Example : Write complex number in polarform

1.Write the following complex numbers in polar form. 2.Write the following complex numbers in
a) 2 + 3i
b) 5 - 2i Try
c) -3 - 7i
This polar form.
Solution: a) 5 - 2i
b) -3 - 7i
a) r = (2)2 + (3)2 = 13 = 3.61
Solution :
 = tan−1 3
2

= 56.3

( )→ 2 + 3i = 3.61 cos56.3 + isin56.3

= 3.6156.3

23

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

Example : Convert complex number to rectangular form

1.Write the following complex numbers in rectangular Try 2.Write the following complex numbers in
form. This rectangular form.

( )6 cos30 + i sin30 2 315

Solution : Solution :

r = 6,  = 30

a + bi = r (cos + isin)

a = 6cos30 = 5.2
b = 6 sin30 = 3

a + bi = 5.2 + 3i

TUTORIAL 5.3 bi
1. Find the modulus and argument from the given Argand diagram below. a
a) b)

bi

a

c) d)
24

bi DUM 10012 Engineering Mathematics 1
a
Unit 5: Complex Numbers

bi

a

2. Find the modulus and the argument of the given complex numbers.

a) z = 8 + 2i b) z = 5 − 3i

c) z = −2 + 3i d) z = −6 − 6i

e) z = 2 + 2i f) z = 5 +12i

25

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

3. Write the following complex number in polar form.

a) 3 +4i b) 5 +4i

c) −12 + 5i d) 1− 2i
e) − 5 − 2i f) 7 −3i
g) 6 + 2i h) − 6 + 5i

i) 4 + 2i j) 6 −3i

4. Write the following complex number in polar form.

26

a) 5(cos32 + isin32) DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

b) 4(cos90 + isin90)

c) 4(cos + isin) d) 3(cos25 + isin25)

e) 8cos + isin   f) 12 cos + isin  

 4 4   2 2 
   

27

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

5. Write the number − 3− −16 as a complex number in a rectangular form.

6. Given Z1 = −2 + 5i and Z2 = −5 − 3i . Find 2Z2 − Z1 .

7. It is given that Z = 1+ 3i . Express Z in the form Z = a + bi.
2 − 5i

8. Write the following complex number Z = −3 − 7i in polar form.

28

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

5.4 DE MOIVRE’S THEOREM

De Moivre’s Theorem is a formula for find power and roots of complex number.

( + ) = (cos + sin )

Example : To evaluate (a+bi)n Try 8

4 This Write (√2 − √2) in the form a + bi
Solution :
Write (√2 − √2) in the form a + bi

Solution : 2

= √ 2 + 2

2

= √(√2) + (√2)

= √4 = 2

Basic angle, = −1 (√2) = 45°

√2

Z1 in quadrant 4. Therefore argument,

= 315°.

4

(√2 − √2)
= [2(cos 4(315°) + sin 4(315°))]4
= 24(cos 1260° + sin 1260°)
= 16(−1 + 0 )
= −16 + 0
= −16

29

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

Example : To evaluate (a+bi)n

Express (1-i) in the form r(cos θ + i sin θ) and Try
hence ( 1- i )12 This

Solution : Write (1 − ) in the form r(cos θ + i sin θ) and
hence ( 1- i )10
= √ 2 + 2 Solution :
= √(1)2 + (−1)2
= √2

Basic angle, = −1 (−1) = 45°

1

Z1 in quadrant 4.

Therefore argument, = 315°.

12

(1 − )12 = (√2 cos 315° + sin 315°)

12 4

= [√2 (cos 12(315°) + sin 4(315°))]

= 64(cos 3780° + sin 3780°)
= 64(−1 + 0 )
= −64

30

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

TUTORIAL 5.4

1. Write (1 + √3 ) in the form r(cos θ + i sin 2. Write (1 + ) in the form r(cos θ + i sin θ) and
hence (1 + )10
3

θ) and hence (1 + √3 )

3. Write (√2 − √2 )in the form r(cos θ + i sin 4. Write (√3 + )in the form r(cos θ + i sin θ) and

θ) and hence 4 7

(√2 − √2 ) hence (√3 + )

31

DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

5. Find the modulus and the argument of the given complex numbers. Hence the z given in form
a+bi

a. (1 − 2 )6 c. (2 + 3 )2

b. (−2 + 3 )5 d. ( 1 + 1 ) 12

√2 √2

32

SUMMARY DUM 10012 Engineering Mathematics 1

Unit 5: Complex Numbers

COMPLEX NUMBER

IMAGINARY ARGAND DIAGRAM
NUMBER bi (imaginary axis)

RECTANGULAR FORM (-ve , +ve) (+ve , +ve)
Real Imaginary (-ve , -ve)
a (real axis)
(+ve , -ve)

MODULUS & ARGUMENT

Modulus, , OPERATION
Argument,
ADDITION
POLAR FORM SUBSTRACTION
OR MUTIPLICATION

CONJUGATE
DIVISION

DE MOIVRE’S THEOREM
( + ) = (cos + sin )

33