Chapter 1 Number System PYQ Answer

Chapter1 : Number System Mathematic
sm015

Q1. 2003/2004 (a) z  i  2  i  1  2i
2i 2i 5 5
Given that the complex number u, v and w such that
z    12   2 2  5
11 1 . If v  1 3i and (b)  5  5 5
u vw w  2  i , express u
 2 
in the form of a  bi , where a and b are real Arg(z) =   tan1 5   2.03rad

numbers. 1
 5 

11 1  1  1 Q3. 2004/2005
u v w 1 3i 2  i Given z1  1 3i and z2  2  5i .

 1  1  3i  1  2  i (a) Express 1  1 in the form of a  bi .
1  3i 1  3i 2  i 2  i z1 z2

 1  3i  2  i (b) Find the argument of z2 in radian
10 5

1  5i
u 10

u  10 (a) 1  1  1  1
5i z1 z2 1  3i 2  5i
 1  1 3i  1  2  5i
 10  5  i 1 3i 1 3i 2  5i 2  5i
5i 5i  1 3i  2  5i
1 9 4  25
 50 10i 49  37 i
26 290 290

 25  5 i (b) z2  2  5i
13 13
Arg( z2 )= tan 1 5   1.190 rad
Q2. 2003/2004 2

Given a complex number z  i Q4. 2004/2005
2i
Given z  x  yi , where x and y are real numbers. If
(a) State z in the form of a  bi z  i  2 , show that 3x2  3y2  8x  6y  7  0
(b) Find the modulus and argument of z
z 1i
.

Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

zi 2 = 3  2i
z 1i 2  i  6i  3i2

(x  yi)  i  2  3  2i   5  5i 
(x  yi) 1 i 5  5i  5  5i 

x  ( y 1)i  2 = 1 1 i
(x 1)  ( y 1)i 2 10

z1  z2   1 2   12  26  0.510
z1z2  2   10  10

x2  ( y  1)2  2 Q6. 2006/2007
(x  1)2  ( y  1)2 Given the complex number z and its conjugate
z satisfy the equation zz  2z  12  6i . Find the
x2  ( y 1)2  2 (x 1)2  ( y 1)2 possible value of z.

Squaring both sides Let z  x  yi , z  x  yi

 x2  ( y 1)2  4 (x 1)2  ( y 1)2 zz  2z  12  6i
 x2  y2  2y 1  4 x2  2x 1 y2  2y 1 (x  yi)(x  yi)  2(x  yi)  12  6i
x2  y2  2x  2yi  12  6i
3x2  3y2  8x  6y  7  0 Comparing
x2  y2  2x  12 ---- (1)
Q5. 2005/2006 2y  6
y  3      (2)
Given two complex number z1  1 3i and Substitute eq(2) into eq(1)
x 2  32  2x  12
z2  2i , express z1  z2 in the form of x2  2x  3  0
z1 z2 (x  3)(x 1)  0
x  3@ x  1
a  bi , where a and b are real numbers.. Hence
determine z1  z2 .

z1 z2

z1  z2  (1  3i)  (2  i)
z1z2 (1  3i)(2  i)

Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

Thus z  3  3i @ z  1  3i z1  5  4  i   5  1  2i 
z2 1  2i 1  2i 

Q7. 2006/2007  4  i   5  10i`
5

An equation in a complex number system is given  4  i 1  2i

1 1  3  3i
 z1
by z  z1 z2   where z1  1 2i and z  9  9  18  3 2

z2  2  i . Find

a) the value of z in the Cartesian form a  bi   tan 1 3   tan 11  
3 4

b) the modulus and the argument of z.

z 3 2 cos    i sin 4`
4  

a) z z1 1 z2   1 Q9. 2008/2009
 z1

 1 1 Given z1 1 i and z2  4  2i . Express z12
(1  2i)  (2  i) 1  2i z1  z2

 1 3i  1 2i in the form of a+bi, where a and b are real
10 5
numbers. [5 marks]
 11i
10 10 . z12   2i   3  3i
z1  z2  3  3i  3  3i
b) | z |  1 2   1 2  50  0.141  6i  6i 2
10  10  50 9  9i 2

Arg(z)=  1   rad  6  6i
tan 1 10  18
 1  4
10  1  1i
33
Q8. 2007/2008
Q10. 2008/2009
If z1  4i and z2  1 2i , find z1  5 . Given that z  x  iy , where x and y are the real
z2 numbers and z is the complex conjugate of z. Find

Hence, express the answer in polar form.

[6 marks]

Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

the positive values of x and y so that 1  2  3  i . 2 = 8 + 6
zz = +
2 = 2 − 2 + 2
[6 marks]

12 1  2 2 − 2 = 8 -------- (1)
z z x  yi x  yi

 3x  yi 2 = 6
x2  y2 = 6 = 3

3x y 2 -------(2)
x2  y2 
  x2 y2 i

3x  3 , x2 y y2 1 2 − ( 3 2 = 8
x2  y2 
)

 3x  3 x2  y2 , x2  y2  y 4 − 8 2 − 9 = 0

x  x2  y2 ( 2 − 9)( 2 + 1) = 0
Then x  y 2 = 9 , 2 + 1= 0

y2  y2  y

2y2  y  0 ∴ = ± 3 , = ± 1

y2 y 1  0 = 3 + , = −3 −

When y  0 , x  0 ( undefined ) (b) When z  3  i
When y  1 , x  1 z  32  12  10

22

Thus

 y 1 , x  1 Arg (z)  tan 1 1   0.322 rad
22 3

Polar form z  10cos0.322  i sin 0.322

When z  3  i

Q11. 2009/2010 z  (3)2  (1)2  10

(a) Given a complex number z  a  bi which Arg(z)    tan 1  1   2.82 rad
satisfy the equation z 2  8  6i . Find all the  3
possible values of .
Polar form z  10cos(2.82)  i sin(2.82)
[6 marks]
Q12. 2010/2011
(b) Hence, epress z in polar form. [4 marks]

(a)
Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

(a) Given two complex numbers z1  2  i and

z2  1 2i . Express z12  1 in the form x  iy , (c) Find z1z2 . Hence, show that z1 z2  z1z2 .
z2 [6 marks]

where x and y are real numbers and z2 is the (a) z1  5  3i and z2  2  i

conjugate of z2 [4 marks]

(b) Given 1  k z1
z1
(b) Hence, find the modulus of z12  1
z2
Method 1
[2 marks]

(a) i) z12  1  2  i 2  1 1  1 2i  1  k(5  3i)
z2  2i 1 2i  5  3i

 3  4i  1 2i  k 1
5 (5  3i)(5  3i)

 15  20i 1 2i 1
5 34

 16 18i Method 2
5
1 1
 16  18 i z1 5  3i
55
 1  5  3i
5  3i 5  3i

(b) z12  1  16 2  18 2  4.82  1 (5  3i)
z2 5 5 34

 1 z1, k  1
34 34

(c)

Q.13 2011/20 z1z2  (5  3i)(2  i)
 10  5i  6i  3i 2
Given two complex numbers z1  5  3i and  13  i
z2  2 i
z1  z2  13  i
(a) State z1 and z2 [1 mark]
z1  z2  (5  3i)(2  i)
(b) Determine the value of k if 1  k z1  10  5i  6i  3i 2
z1  13  i

z1  z2  z1  z2

[3 marks]

Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

Q14. 2012/2013 (a) p  q  1 5 i
4  2i 4  2i 2
Given a complex number z 1 3i .
p  4  2i  q  4  2i  1 5 i
Determine the value of k if z2  k 1  4  2i 4  2i 4  2i 4  2i 2
z
4 p  2 pi  4q  2qi  1 5 i
. 16  4 16  4 2
[7 marks]

[ 7 marks] 4 p  4q  2 p  2q  1 5 i

Method 1 20 20 2

Comparing

z 2  2  2 3i

k  1   k  1  1  3i   k  k 3i 4( p  q)  1
 z   1  3i 4 20 .........(1)
1 3i
pq5

 2  2 3i  k  k 3i 2( p  q)  5
44 20 2 ------- (2)

 k  2  k  8 p  q  25
4

Method 2 Solve eq (1) & eq (2)
p  15 & q  10
kz z 2  (1  3i)(2  2 3i)
 (2  2 3i  2 3i  6i 2 )   (b) 143x1 82x3  7
8    23x1 73x1 26x9  7

Q15. 2012/2013

(a) Find the values of p and q if 29x10  73x
log 29x10  log 73x
p  q 1 5i
4  2i 4  2i 2 9x 10log 2  3xlog 7

(b) Given log10 2  m and log10 7  n . Express x (9x  10)m  3xn
in term of m and n if (9m  3n)x  10m

  143x1 82x3  7

Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

x  10m   10m z has a pure imaginary part only.
9m  3n 3(3m  n)
(c) z(2  i)  (z 1)(1 i)
Q16. 2014/2015
Let z  a  bi be a non-zero complex number (a  bi)(2  i)  (a  bi  1)(1  i)
(a) Show that 1  z
2a  ai  2bi  b  a  ai  bi  b  1  i
z z2 2a  (2b  a)i  (a  1)  (a  b  1)i
(b) Show that if z  z , then z is a complex
By comparing
number with only an imaginary part. 2a  a 1
(c) Find the value of a and b if a 1
2b  a  a  b 1
z(2  i)  (z 1)(1 i) . 3b  3
b 1

Q17. 2015/2016

(a) 1  a 1   a  bi   a  bi Given z1  3  3i and z2  3  2i .
z  bi  a  bi  a2  b2

z  a2  b2 (a) Write z1 in polar form.

z 2  a2  b2 (b) Express z1 z 2  i3 in the form a  bi ,
13  z2

z  a  bi a,b  R
z2 a2  b2

Therefore, 1  z shown. (a) z1  3  3i
z z2 z1  32  (3)2  18  3 2

(b) Given z  z Arg (z)  tan 1 3   
3 4
a  bi  (a  bi)
Polar form z1  3 2 cos    i sin   
a  bi  a  bi  4   4 
2a  0
a0 (b) z1z2  (3  3i)(3  2i)  3  15 i
13 13 15 13
z  a  bi  0  bi  bi
Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

i3   i Q19. 2017/2018
 z2  (3  2i)
Given a Complex number z  2  i
 i  3  2i
3  2i 3  2i (a) Express z  1 in the form a  bi , where a and b are
z
= 2  3i
13 13 real number.

i3  2  3 i (b) Obtain the z  1 . Hence, determine the values of
 z2 13 13 z

Therefore z1 z 2  i3   3  15 i    2  3 i  real numbers  and  if
13  z2 13 13  13 13 
  i  z  1  z  1 2
 5  12 i z  z
13 13
(a) z  1  (2  i)  1`
Q18. 2016/2017 z 2i
 (2  i)  1  2  i
Solve for p and q where p  q , such that 2i 2i
 (2  i)  (2  i)
 ( p  qi)  3  16  i3 4 1
3i 2i 2  i
55
 8  6i
55

16  16  1  4i (b)  8 2    6 2  2
i3  i2  i  i z1  5  5 
z
 ( p  qi)  3  16  i3
3i   i  z  1  z  1 2
z  z
( p  qi)  3  4i  i
 285  6 i2
3i 5
( p  qi)  3i(3  5i)
 2 64  96 i  36 
p  qi  15  9i 25 25 25 
 p  15, q  9
 56  192 i   56 ,   192
Edited by limhweecheng 25 25 25 25

Chapter1 : Number System Mathematic
sm015

  i  z  1  z  1 2 Q21. 2019/2020
z  z
Given the complex numbers z1  i and
2 8 6 i 2
5 5 z2  2  i 3
 
(a) Express z12 and z2 in the from a  bi ,
 2 64  96 i  36  where a, b   .
25 25 25 
(b) From part 1(a), find W  z12  z2 . Hence, find
 56  192 i   56 ,   192 z1
25 25 25 25
W and arg W
(c)

Q20. 2018/2019  (a) z12  (i)2   1 2  1

Given z1  2  3i and z2  4  4i . Express z2  i3 z2  2  i 3
z1  z2
(b) W  z12  z2  1  (2  i 3)
in Cartesian form. z1  i
 1i 3
z2  i3  4  4i  i3 , i3  i i
z1  z2 2  3i  (4  4i)
 1i 3  i
 4  4i  (i) i i
2  3i  (4  4i)
 i  3i 2
 4  4i  i  i2
2  3i 4  4i
 i  3(1)
 4  4i   2  3i   i   4  4i   (1)
2  3i  2  3i  4  4i  4  4i 
 3i

 20  4i   4  4i W  ( 3)2  12  2
4  9 16 16

 20  4 i  1  1 i Arg W = tan 1  1   
13 `13 8 8 3 6

 147  45 i
104 104

Edited by limhweecheng

Chapter1 : Number System Mathematic
sm015

 1i 3  i
i i

 i  3i 2
 i2

 i  3(1)
 (1)

 3i

W  ( 3)2  12  2

Arg W = tan 1  1   
3 6

Edited by limhweecheng