10
COMPLEX NUMBERS
Theory and Illustrations
COMPLEX NUMBERS
[2]
COMPLEX NUMBERS
In our daily mathematics, we deal with numbers like 2, 3, ........ These numbers are all known as real numbers
and can be shown on real number line.
Till now, the numbers which we came across in mathematics are all real numbers.
Now, in this chapter we will study complex numbers. These complex numbers cannot be shown on real number
line neither practically. You can comprehend its application. But nevertheless complex numbers plays vital role
as a very important tool in solving varieties of problems.
Therefore, understanding complex numbers is very important in development of higher mathematics.
IOTA (i) : i = (–1)
The basis of all complex number theory is concept of ‘iota’, i = (–1).
We know, that inside square root quantity should not be negative, such a quantity [say (–5)] has no physical or
real meaning.
In complex number we define a quantity called ‘iota (i)’ where i = (–1). Since ‘i’ is not real but imaginary,
wherever there is ‘i’ that number becomes complex number and can be studied using theory of complex numbers
NOTE:
(a) Although, in real number, (–1) has no meaning and cannot be studied but in complex number we will study
even square root of negative numbers using i = (–1), (iota).
(b) This complex number has property which are very-very useful in solving varieties of problems belonging to
all kinds of fields.
Illustration 01:
Write the following in terms of ‘i’:
(a) (–5)
(b) (–16)
SOLUTION:
(a)
(–5) = 5 · (–1) = 5i
(b)
(–16) = 16) · (–1) = 4i
(c)
(–121) = 121) · (–1) = 11i
(c) (–121)
{ i = (–1)}
Thus, we can write square root of any negative number in terms of IOTA ‘i’.
COMPLEX NUMBERS

[3]
PROPERTIES OF i :
1.
i = (–1)
2.
i2 = –1
3.
i3 = i2·i = –i
4.
i4 = (i2)2 = (–1)2 = +1
Thus, every fourth root of ‘i’ becomes 1 and so any high power can be reduced to lower power.
(a) i4n = (i4)n = 1
e.g.:
(b) i4n + 1 = (i4)n·i = i
(c) i4n + 2 = (i4)n·i2 = i2
(d) i4n + 3 = (i4)n·i3 = i3 = –i
5.
1 i
  i
i i2
Illustration 02:
Find the value of:
(a) i453
SOLUTION:
(b) 1 + i2 + i4 + ......... i2n
(c) i14 + i120 + i333 + i403
We know, i4 = 1, so try to write the power of ‘i’ as multiple of 4.
(a)
i453 = (i4)113 × i = (1)113 × i = i
(b)
1 + i2 + i4 + ......... i2n
=
(c)
1[(i 2 )n1  1]
i2 1
{ i = (–1)}
=
(1)n1  1
1  1
If n is odd then
(1)even  1 1  1

0
2
2
If n is even then
(1)odd  1 1  1

 1
2
2
i14 + i120 + i333 + i403
= (i4)3·i2 + (i4)30 + (i4)83·i + (i4)100·i3
= (1)3·i2 + (1)30 + (1)83·i + (1)100·i3
= i2 + 1 + i + i3
= 1 + i + (–1) + (–i)
=0
COMPLEX NUMBERS

[4]
GENERAL COMPLEX NUMBER:
A general complex number is denoted by ‘z’ and is given by
z = x + iy or z = a + ib
Let us consider, z = x + iy
where x is real part of complex number ‘z’ and y is imaginary part of complex number ‘z’.
(a) If y = 0 then z = x is purely real number.
 Thus, all real numbers can be seen as complex number with imaginary part zero or real numbers are
subset of complex numbers.
e.g.:
2 is a real number and can be written as 2 + i·0 (complex number)
or
–3 = –3 + i·0
(complex number notation)
(b) If x = 0 then z = iy is purely imaginary
(c) i = (–1)
In general, z = x + iy contains real part and imaginary part.

PROPERTIES OF COMPLEX NUMBER:
1.
(x1 + iy1) + (x2 + iy2) = (x1 + x2) + i(y1 + y2)
i.e., real part is added separately and imaginary part separately.
2.
(x1 + iy1) – (x2 + iy2) = (x1 – x2) + i(y1 – y2)
3.
(x1 + iy1)(x2 + iy2)
= x1x2 + ix1y2 + iy1x2 + i2y1y2
= (x1x2 – y1y2) + i(x1y2 + x2y1)
whenever we multiply two or more complex numbers, we should first multiply in usual way and then
susbtitute i2 = –1 and separate real part and imaginary part separately.
4.
x1  iy1
x2  iy2
(Division of two complex numbers)
=
=
=
5.
x1  iy1 x2  iy2

(multiplying numerator and denominator by conjugate of denominator)
x2  iy2 x2  iy2
x1x2  ix1 y2  iy1x2  i 2 y1 y2
x22  i 2 y22
( x1x2  y1 y2 )  ( y1x2  x1 y2 )
x22  y22
0 + 0·i is known as zero complex number
{ i2 = –1}
COMPLEX NUMBERS
6.
[5]
Two complex numbers are said to be equal if their real part and imaginary parts separately equal.
i.e,
If x1 + iy1 = x2 + iy2 implies that x1 = x2 and y1 = y2
e.g.: If x + iy = 2 then find x and y.
x + iy = 2 + i·0
Equating real and imaginary part separately both sides, we get
x = 2 and y = 0
7.
We cannot compare two complex numbers that is if we want to find which is greater among 2 + 3i and
3 – 4i, we cannot say anything.
Illustration 03:
Write the following complex numbers in the form of A + iB
(a) (2 + 3i) – (3 – 5i)
(e)
(b) (1 + 2i)2
(c) (1 – i)3
3  4i
5  2i
SOLUTION:
(a)
(2 + 3i) – (3 – 5i) = (2 – 3) + i(3 + 5)
= –1 + 8i
(b)
(1 + 2i)2 = 12 + (2i)2 + 2·1·(2i)
= 1 + 4i2 + 4i
= 1 – 4 + 4i
= –3 + 4i
(c)
(1 – i)3 = 13 + (–i)3 + 3(1)2(–i) + 3(1)(–i)2
= 1 – i3 – 3i + 3i2
= 1 – (–i) – 3i – 3
= –2 – 2i
(d)
(2 + 3i)(1 – i) = 2 – 2i + 3i – 3i2
= 2 + i – 3(–1)
=5+i
(e)
3  4i 3  4i 5  2i

=
5  2i 5  2i 5  2i
=
15  6i  20i  8i 2
52  22 i 2
=
7  26i
7  26i
=
25  4
29
=
7 26
 i
29 29
(d) (2 + 3i)(1 – i)
COMPLEX NUMBERS
[6]
Illustration 04:
If (x + y) + i(x – y) = 2 + 3i then find the values of x and y.
SOLUTION:
Given,
(x + y) + i(x – y) = 2 + 3i
So,
x+y=2
....... (1)
and
x–y=3
....... (2)
Adding (1) and (2), we get
2x = 5  x = 5 2
Now, subtracting (2) from (1), we get
2y = –1  y = 1 2
Illustration 05:
If 4x + i(3x – y) = 3 – 6i where x, y  R, then find x and y.
SOLUTION:
Comparing the given equation, 4x + i(3x – y) = 3 – 6i
4x = 3 and 3x – y = –6
 x = 3 4 and y = 3x + 6 =
3 3
33
6
4
4
Illustration 06:
If z = 2 – 3i, show that z2 – 4z + 13 = 0.
SOLUTION:
L.H.S. = z2 – 4z + 13
= (2 – 3i)2 – 4(2 – 3i) + 13
= 4 + 9i2 – 12i – 8 +12i + 13
= 4 – 9 – 8 + 13
= 17 – 17 = 0 = R.H.S.
Illustration 07:
Solve for x, y (x, y  R) if (1 + i)y2 + (6 + i) = (2 + i)x
SOLUTION:
In such questions, express L.H.S. and R.H.S. in the standard form x + iy
(1 + i)y2 + (6 + i) = (2 + i)x
 (y2 + 6) + i(y2 + 1) = 2x + ix
Now equating real and imaginary parts:
y2 + 6 = 2x and y2 + 1 = x
Solving for x and y: x = 5 and y = ±2
COMPLEX NUMBERS
[7]
Illustration 08:
Prove that x4 + 4 = (x + 1 + i)(x + 1 – i)(x – 1 + i)(x – 1 – i)
SOLUTION:
Consider R.H.S. = (x + 1 + i)(x + 1 – i)(x – 1 + i)(x – 1 – i)
= [(x + 1)2 – i2][(x – 1)2 – i2]
= [x2 + 2x + 2][x2 – 2x + 2]
= (x2 + 2)2 – (2x)2 = x4 + 4 + 4x2 – 4x2
= x4 + 4 = L.H.S.
Illustration 09:
If (a + ib)2 = x + iy, then find the value of x2 + y2.
(a + ib)2 = x + iy
SOLUTION:
 a2 + 2abi + i2b2 = a2 – b2 + 2abi = x + iy
 x = a2 – b2, y = 2ab
 x2 = (a2 – b2)2, y = 4a2b2
 x2 + y2 = (a2 – b2)2 + 4a2b2
 x2 + y2 = a4 + b4 – 2a2b2 + 4a2b2 = a4 + 2a2b2 + b4
 x2 + y2 = (a2 + b2)2

CONJUGATE COMPLEX NUMBERS:
If z = x + iy then its complex conjugate is a denoted by z and is equal to x – iy i.e., z  x  iy , where z is
complex conjugate of z.
e.g.: If z = –2 – 3i then find the conjugate of z.
If z = –2 – 3i then complex conjugate will be z  2  3i .

PROPERTIES OF CONJUGATE COMPLEX NUMBERS:
Let z = x + iy
1.
z  z  2 x  2 Re( z )
{purely real number}
2.
3.
z  z  2iy  2 Im( z )
{purely imaginary number}
{Remember}
z  z = (x + iy)(x – iy) = x2 – i2y2 = x2 + y2 > 0
Thus, z  z is always positive.
4.
z1  z2  z1  z2 Conjugate of sum = Sum of conjguates
5.
z1  z2  z1  z2
6.
z n  ( z )n
7.
( z )  z Conjugate of conjugate is original complex number..
8.
 z1  z1
 
 z2  z 2
Conjugate of product = Product of conjugates
COMPLEX NUMBERS
[8]
Illustration 10:
Write the conjugate of the complex numbers:
(a) 3 + 4i
SOLUTION:
(b) (15 + 3i) – (4 – 20i)
(a)
If z = 3 + 4i then conjugate is 3 – 4i
(b)
First simplify:
(c) 
(15 + 3i) – (4 – 20i)
= 11 + 23i, then conjugate is 11 – 23i
(c)
First simplify:

5i
7i
=
=
5i 7  i

7 i 7 i
35i  5i 2
72  i 2
=
35i  5
49  1
=
5  35i
1  7i
or
50
10
=
1 7
 i then its conjugate will be  1  7 i
10 10
10 10
Illustration 11:
Find the conjugate of
SOLUTION:
1 i
(1  i )(2  3i )
First bring it to A + iB form then conjugate is given by A – iB.
1 i
(1  i)(2  3i)
=
=
=
1 i
2  3i  2i  3i 2
1 i 5  i

5i 5i
5  i  5i  i 2
52  12
=
6  4i
26
=
6 4
 i
26 26
 The conjugate is given by
6 4
 i
26 26
5i
7i
COMPLEX NUMBERS
[9]
Illustration 12:
If z = 3 – 4i then find z  z
SOLUTION:
We know, z  z = x2 + y2
= 32 + (–4)2
= 9 + 16
= 25
Illustration 13:
If z = 2 + 3i then find the value of (z – 2)( z + 1)
SOLUTION:
(z – 2)( z + 1)
 (2 + 3i – 2)(2 – 3i + 1)
 (3i)(1 – 3i)
 3i – 9i2
 9 + 3i
Illustration 14:
Find the real and imaginary parts of
SOLUTION:
2 i
(2  3i )(1  i )
We have to simplify like discussed earlier.
2i
(2  3i)(1  i)
=
=
=
2i
2  2i  3i  3i 2
2i 5i

5i 5i
10  2i  5i  i 2
52  12
=
11  3i
26
=
11 3
 i
26 26
Real part is
11
3
and Imaginary part is .
26
26
Don’t do mistake saying real part is 11 and imaginary part is 3.
COMPLEX NUMBERS
[10]
Illustration 15:
Find the real and imaginary part of
SOLUTION:
1
1  cos 2  i sin 2
We have to bring it in A + iB form. We know, whenever complex number is in denominator we
have to multiply with conjugate up and down.



(1  cos 2)  i sin 2
(1  cos 2)2  sin 2 2
2cos2   i  2sin  cos 
1  cos2 2  2cos 2  sin 2 2

2cos2   i  2sin  cos 
2  2cos 2

cos2   i  sin  cos 
1  cos 2

Real part is
1
(1  cos 2)  i sin 2

(1  cos 2)  i sin 2 (1  cos 2)  i sin 2
cos2   i  sin  cos 
2cos2 

cos   i  sin 
2cos 

1
1
 i  tan 
2
2
1
tanθ
and Imaginary part is 
2
2
Illustration 16:
Find the additive inverse of 3 – 4i.
SOLUTION:
Additive inverse of a complex number z is given by z’ such that z + z’ = 0
So,
(3 – 4i) + z’ = 0
 z’ = –3 + 4i

–3 + 4i is the additive inverse of 3 – 4i.
COMPLEX NUMBERS
[11]
Illustration 17:
Find the multiplicative inverse of 3 – 4i.
SOLUTION:
Multiplicative inverse of a complex number z is given by z’ such that z·z’ = 1
So,
(3 – 4i)·z’ = 1
 z’ =
1
3  4i
 z’ =
3  4i
(3  4i)(3  4i)
 z’ =

3  4i
92  42
 z’ =
3  4i
25
 z’ =
3 4
 i
25 25
GEOMETRICAL REPRESENTATION OF COMPLEX NUMBERS:
A complex number z = x + iy can be represented by a point (x, y) on the co-ordinate plane. This plane is called as
Argand plane.
Thus, any complex number z = x + iy is represented as a point (x, y) where x is real part and y is imaginary part.
Argand Plane
y or Im(z)
P(x, y) or x + iy
x
x or Re(z)
O
y
Illustration 18:
Plot the following complex numbers on the Argand plane:
(a) z = 2 + 3i
SOLUTION:
(b) z = 2 – 3i
(a)
(c) z = –2 + 3i
(d) z = –2 – 3i
2 + 3i has to be plotted as (2, 3). Real part x = 2, imaginary part y = 3
y
x
O
y
2
(2, 3)
3
x
COMPLEX NUMBERS
[12]
(b)
2 – 3i has to be plotted as (2, –3). Real part x = 2, imaginary part y = –3
y
2
x
O
x
3
(2, 3)
y
(c)
–2 + 3i : Real part x = –2, imaginary part y = 3
y
(2, 3)
3
x
2
x
O
y
(d)
–2 – 3i : Real part x = –2, imaginary part y = –3
y
2
x
3
x
O
(2, 3)
y
Illustration 19:
Write the complex number represented by points P and Q on Argand plane:
y
Q(3, 4)
x
P(2, 3)
O
x
y
SOLUTION:
On argand plane, point (x, y) is basically complex number z = x + iy.
Similarly,
P(2, 3) represents 2 + 3i
and
Q(–3, 4) represents –3 + 4i
COMPLEX NUMBERS

[13]
MODULUS OF A COMPLEX NUMBER:
We studied that a complex number z = x + iy can be represented by a point (x, y) on the argand plane as shown:
y
P(x, y)
x
r

x
O
y
x
y
OP is called modulus of complex number z = x + iy.
Modulus is represented by | z | or r and given by | z | = x2  y 2 = Re( z )2  Im( z )2
NOTE:
1.
Modulus of a complex number is always a positive real number.
2.
It represents distance of z from origin.
3.
If z = x + iy then x2 + y2 = | z |2.
Illustration 20:
Find the modulus of the following complex numbers:
(a) 2 + 3i
SOLUTION:

(b) –2 + 3i
We know, modulus of complex number is given by x2  y2
(a)
|2 + 3i| = 22  32 = 4  9 = 13
(b)
|–2 + 3i| = (2)2  32 =
4  9 = 13
PROPERTIES OF MODULUS OF A COMPLEX NUMBER:
1.
z  z = x2 + y2 = | z |2;
If | z | = 1  z  z = 1 or z 
1
z
Very Important Property: This property allows us to get rid of modulus.
e.g.:
| 2 + 3i |2 = (2 + 3i) (2  3i) = (2 + 3i)(2 – 3i)
Thus, we are free from modulus.
2.
| z | = | z | = | –z | . In each case modulus will be x2  y2
3.
| z1·z2 | = | z1 | · | z2 |
4.
z1 | z1 |

z2 | z2 |
5.
| zn | = | z |n
COMPLEX NUMBERS
[14]
Illustration 21:
Find the modulus of the following complex numbers:
(a) 2 – 3i
(e)
(c) (2 – i)2
(b) (2 + 3i)(1 – i)
(d)
(1  i )(2  i )2
(3  4i )3 (5  i )
1
1  cos   i sin 
SOLUTION:
In finding the modulus we should not first simplify expression to A + iB and then find modulus.
Rather we should use properties given on modulus.
(a)
Modulus of 2 – 3i is
| 2 – 3i | = 22  (3)2  4  9  13
(b)
|(2 + 3i)(1 – i)|
= | 2 + 3i | | 1 – i |
2
2
2
2
= ( 2  3 )( 1  (1)  13  2  26
(c)
|(2 – i)2|
= | 2 – i |2
2
2 2
2
= ( 4  (1) )  ( 5)  5
(d)
(1  i)(2  i)2
(3  4i)3 (5  i )
=
=
(e)
1
1  cos   i sin 
|1  i | | 2  i |2
| 3  4i |3 | 5  i |
=
( 2)  ( 5)2
(5)3 ( 26)
1
2
=
25  2  13 25 13
1
= |1  cos   i sin  |
=
=
=
=
=
1
(1  cos )2  sin 2 
1
2
1  cos   2cos   sin 2 
1
=
2  2cos 
1
2 2sin 2 
=
2
1
2 2 | sin  |
2
1
2(1  cos )
1
2 2 sin 2 
2
{ x2  | x | }
COMPLEX NUMBERS
[15]
Illustration 22:
Find x2 + y2 in each case:
(a)
(2  i )
= x + iy
(3  i )(4  i )
SOLUTION:
(b) (2 + 3i)3 = x + iy
(c) (2 + 3i)(3 – i) = x + iy
We know for complex number x + iy the quantity x2 + y2 represents square of modulus. So in
each case above we have to find square of modulus of complex number.
(a)
x2 + y2 =
|x + iy|2
2i
=
(3  i)(4  i)
2
 | 2i| 
=

 | 3i | | 4 i | 

5 

= 
 10 17 
2
2
= 1 34
(b)
x2 + y2 = |x + iy|2 = |(2 + 3i)3|2
= (|2 + 3i|3)2
= |2 + 3i|6
= ( 13)6
= 133 = 2197
(c)
x2 + y2 = |x + iy|2 = |(2 + 3i)(3 – i)|2
= |2 + 3i|2 |3 – i|2
= ( 13)2 ( 10)2
= 130
Illustration 23:
If (a1 + ib1)(a2 + ib2) = x + iy then find the value of (a12 + b12)(a22 + b22)
SOLUTION:
Given
(a1 + ib1)(a2 + ib2) = x + iy
........ (1)
Taking square of modulus both sides
|a1 + ib1|2 · |a2 + ib2|2 = |x + iy|2
 (a12 + b12)(a22 + b22) = x2 + y2
........ (2)
COMPLEX NUMBERS

[16]
ARGUMENT OR AMPLITUDE OF A COMPLEX NUMBER:
We studied previously, a complex number z = x + iy is denoted by a point P(x, y) on the argand plane where
y
r=
x
O
|z
| P(x, y)

x
y
x
y
— OP (modulus) = | z | = x2  y 2
y
x
— Argument of z is angle  which this line OP makes with the positive x-axis thus, tan   
Im z
Re z
To understand argument or amplitude of complex number ‘z’ we have to understand it more clearly!
1.
Argument of complex number ‘z’ is given by angle ‘’, but the angle ‘’ must belong to (–, ]
That is in complex number the value of ‘’ does not change from 0 to 2 as in trigonometric plane but goes
from – to , including  but not –
2.
The way we read angle ‘’ is as shown.
/2
0 to 


0
0 to 
/2
If ‘z’ is in upper half then ‘’ will be positive and varies from 0 to . If ‘z’ is in lower half then ‘’ will be
negative and varies from 0 to –.
3.
The value of ‘’ that we get between (–, ] is called principal argument and general argument is given by
 + 2k, k  I.
4.
Thus to find argument we have to first locate where ‘z’ is located and then find ‘’. Below illustrations will
clarify and should be studied properly.
5.
The value of argument should be in general given in radians.
COMPLEX NUMBERS
[17]
Illustration 24:
Find the principal argument and general arguments for the following complex numbers:
(a) 1 + 3i
SOLUTION:
(b) 1 – 3i
(c) –1 – 3i
(d) –1 + 3i
To find argument we have to show the complex number on argand plane and see the angle
which the line ‘OP’ makes with positive x-axis and whether angle is positive or negative. To
clarify see the solving below:
(a)
z = 1 + 3i
y
P(1, 3)
x
O

1
3
x
y
tan  =
Magnitude of ‘’:
3
= 3
1
 = 60o =  3
Sign of ‘’:
Since, the complex number lies in upper quadrant, it will be positive
So,
Principal argument =  3
General argument =  3 + 2k
(b)
z = 1 – 3i
y
x
O
1

x
3
P(1, 3)
y
Magnitude of ‘’:
tan  =
3
= 3
1
 = 60o =  3
Sign of ‘’:
Since, the complex number belongs to lower quadrant, it is negative
So,
Principal argument = –  3
General argument = –  3 + 2k
COMPLEX NUMBERS
[18]
(c)
z = –1 – 3i
y
1
x
3
O


x
P(1, 3)
y
Magnitude of ‘’:
Angle ‘’ is shown on graph, but we can find first  and then ‘’
tan  =
3
= 3
1
 = 60o =  3
 =  –  3 = 2 3
Sign of ‘’:
Since, the complex number belongs to lower quadrant, it will be
negative.
Principal argument = – 2 3
So,
General argument = 2k – 2 3
(d)
z = –1 + 3i
y
P(1, 3)
3
x


O
1
x
y
Magnitude of ‘’:
First we find 
tan  =
3
= 3
1
 = 60o =  3
 =  –  =  –  3 = 2 3
Sign of ‘’:
Since, the complex number ‘z’ is in above quadrant, it will be
positive.
So,
Principal argument = 2 3
General argument = 2k + 2 3
COMPLEX NUMBERS
[19]
Illustration 25:
Find the argument of following complex numbers:
(a) 2
(b) 2i
(c) –2
(d) –2i
(e) 1 + i
SOLUTION:
Like we did in previous illustration first we have to locate the complex number ‘z’ and then find
the magnitude and sign of argument of complex number ‘z’.
(a)
z = 2 = 2 + 0i
y
x
P(2, 0)
x
2
O
y
Magnitude of ‘’:
Angle ‘’ = 0
Sign of ‘’:
None
So,
Principal argument = 0
General argument = 2k + 0 = 2k
(b)
z = 2i = 0 + 2i
y
2 P(0, 2)
x

O
x
y
Magnitude of ‘’:
Angle ‘’ =  2
Sign of ‘’:
Since, complex number ‘z’ is in upper half, argument will be positive
So,
Principal argument =  2
General argument =  2 + 2k
(c)
z = –2 = –2 + 0i
y
P(2, 0)
x
2

O
y
x
COMPLEX NUMBERS
[20]
Magnitude of ‘’:
From figure it is clear that magnitude is .
Sign of ‘’:
When complex number is on negative x-axis we take argument +
and not –.
Principal argument = 
So,
General argument =  + 2k
(d)
z = –2i = 0 – 2i
y
x
O 
2
P(0, 2)
x
y
Magnitude of ‘’:
It is clear that, ‘’ =  2
Sign of ‘’:
Since, complex number ‘z’ is in lower half, argument will be negative
Principal argument = –  2
So,
General argument = –  2 + 2k
(e)
z=1+i
y
x
O
P(1, 1)
1

x
1
y
Magnitude of ‘’:
tan  =
1
=1
1
  = 4
Sign of ‘’:
Since, complex number ‘z’ is in upper quadrant, argument will be
positive.
So,
Principal argument =  4
General argument = 2k +  4
COMPLEX NUMBERS

[21]
PROPERTIES OF ARGUMENT:
1.
arg(z1·z2) = arg(z1) + arg(z2)
2.
z 
arg  1  = arg(z1) – arg(z2)
 z2 
3.
arg(zn) = n arg(z)
4.
arg(z1·z2 ........... zn) = arg(z1) + arg(z2) + ........... + arg(zn)
5.
arg( z ) = –arg(z)
Illustration 26:
Find the modulus and argument of the following complex numbers:
(a) (1 + i)5
SOLUTION:
(b)
(a)
1  3i
1 i
(1 + i)5
Modulus:
|(1 + i)5| = |1 + i|5
= ( 12  12 )5 = (2)5 = 42
Argument:
arg(1 + i)5 = 5·arg(1 + i)
y
Now, arg(1 + i)
tan  =
1
=1
1
x
O
  = 4
arg(1 + i)5 = 5·arg(1 + i) = 5· 
(b)
4
= 5
P(1, 1)
1

x
1
y
4
1  3i
1 i
Modulus:
1  3i |1  3i |

1 i
|1  i |
=
Argument:
1 3 2

= 2
11
2
 1  3i 
arg 
 = arg(1 + 3i) – arg(1– i)
 1 i 
y
= x
O
y
P(1, 3)
 3
1
x
x
y
=  3 – (–  4 ) =
O
1

y
4 3 7

12
12
x
1
P(1, 1)
COMPLEX NUMBERS
[22]
Illustration 27:
If π 6 and π 3 are the arguments of z1 and z2 then find the value of arg(z1·z2)
SOLUTION:
We know that
arg(z1·z2) = arg(z1) + arg(z2)
= 6  3
=
  2 

6
6
= 2
Illustration 28:
z 
 z2 
If π 4 and π 3 are the arguments of z1 and z2 then find the value of arg  1 
SOLUTION:
We know that
z 
arg  1  = arg(z1) – arg(z2)
 z2 
Now, given
arg( z1 ) =  4
 –arg(z1) =  4
and
 arg(z1) = –  4
....... (1)
arg(z2) =  3
....... (2)
Substituting (1) and (2), we get
z 
arg  1  = arg(z1) – arg(z2)
 z2 
= –  4 – 3
=  712
COMPLEX NUMBERS

[23]
TRIGONOMETRIC(POLAR) AND EXPONENTIAL(EULER’S) REPRESENTATION:
We studied, that any complex number z = x + iy is a point P(x, y) on argand plane as shown.
y
x
r=
O
|z
| P(x, y)

x
y
x
y
This representation of z = x + iy is called cartesian representation.
There is another way of representing complex number ‘z’ in terms of modulus (| z | = r) and principal
argument ().
This representation in ‘r’ and ‘’ is called Polar or Euler’s representation.
x = r cos  and y = r sin 
Now, from above:
So,
z = x + iy
 z = r cos  + i r sin 
 z  r (cos   i sin ) is Trigonometric form or Polar form
where r is modulus(must be positive) and  is principal argument
Euler gave derivation proving
So,
ei = cos  + i sin 
z  r  ei is Exponential or Euler’s reprensentation
This is very useful form.
Thus, to write trigonometric or exponential representation we need to find modulus ‘r’ and principal
argument ‘’. Then
z = r·ei = r(cos  + i sin )
COMPLEX NUMBERS
[24]
Illustration 29:
Write the following in trigonometric and exponential form:
(a) z = 1 + i
(b) z = –3 + i
SOLUTION:
To write a complex number in trigonometric or exponential form, we need to find modulus ‘r’
and principal argument ‘’.
(a) z = 1 + i
Modulus:
r = | z | = | 1 + i | = 12  12 = 2
y
Principal Argument:
x
P(1, 1)
1

x
1
O
y
tan  =
1
= 1   = 4
1
Trigonometric form: z = r(cos  + i sin )
 z = 2(cos  4 + i sin  4 )
i
Exponential form: z = rei  z = 2 e
4
(a) z = –3 + i
Modulus:
Principal Argument:
r = | z | = | –3 + i | = 3  1 = 2
y
P(3, 1)
1
x

3

O
x
y
tan  =
1
  = 6
3
 =  –  = 5 6
(above half so positive)
Trigonometric form: z = r(cos  + i sin )
 z = 2(cos 5 6 + i sin 5 6 )
i5
Exponential form: z = rei  z = 2 e
6
COMPLEX NUMBERS

[25]
IMPORTANT NOTE:
1.
ei can be written as 1·ei i.e., it is the complex number where modulus is 1 and principal argument is ‘’.
Thus, |ei| = 1.
2.
ei = cos  + i sin 
3.
e–i = cos(–) + i sin(–)  ei  cos   i sin 
4.
Write 1, i, –1, –i in polar form.
We see modulus of all of them is 1.
y
1 = 1·ei 0
i
x
i = 1·ei·/2
1
O
1
x
i
1 = 1·ei·
i = 1·ei·
y
Illustration 30:
Prove that
(cos 1 + i sin 1)(cos 2 + i sin 2) ....... (cos n + i sin n) = cos(1 + 2 ....... + n) + i sin (1 + 2 ....... + n)
SOLUTION:
Most of such problems can be done by using exponential form.
L.H.S. = (cos 1 + i sin 1)(cos 2 + i sin 2) ....... (cos n + i sin n)
= ei·1 · ei·2 · ei·3 .......... ei·n
= e(i·1 + i·2 + i·3 .......... + i·n)
= ei(1 + 2 + 3 .......... + n)
= cos(1 + 2 + ......... + n) + i sin(1 + 2 + ......... + n)
= R.H.S.
Hence Proved
See how powerful is Exponential form
COMPLEX NUMBERS
[26]
Illustration 31:
Simplify: (a) (1 + 3i)5
SOLUTION:
(b) (1 – i)7
Exponential form is very easy and useful in complex numbers. Thus, whenever we have powers
or want to do some operation, we should keep exponential form for different complex numbers.
(a) (1 + 3i)5
y
Consider z = 1 + 3i
Modulus:
P(1, 3)
| z | = 1 3 = 2
x
3
Principal argument: tan  =
= 3
1
O
  = 60o =  3
i
Exponential form: 1 + 3i = 2· e
So,
i5
(1 + 3i)5 = 25· e

1
3
x
y
3
i5
3 = 32· e
3
= 32(cos 5 3 + i sin 5 3 )
= 32{cos (2   3 ) + i sin (2   3 ) }
= 32(cos  3 – i sin  3 )
= 32 cos  3 – 32i sin  3
= 16 – 163i
(b) (1 – i)7
Consider z = 1 – i
Modulus:
y
| z | = 1 1 = 2
1
=1
1
Principal argument: tan  =
x
O
  = – 4
Exponential form: 1 – i = 2· e
So,
(1 – i)7 = (2· e
i
y
i
4
4 )7 = 27/2· e
i7 
4
= 27/2{cos ( 7 4) + i sin ( 7 4) }
= 82(cos 7 4 – i sin 7 4 )
= 82{cos (2   4) – i sin (2   4) }
= 82(cos  4 + i sin  4 )
= 82( 1 2  i 2 )
= 8 + 8i
1

x
1
P(1, 1)
COMPLEX NUMBERS
[27]
Illustration 32:
Write the following in trigonometric and exponential forms:
(a) (1 – i)5
(b) cos π 3 – i sin π 3
(c) –cos π 3 + i sin π 3
(d) 2(sin π 3 + i cos π 3 )
(e) –2(cos π 6 + i sin π 6 )
(f) 2(cos 300o – i sin 300o)
(g) 1 + cos  8π   + i sin 8π
(h)

SOLUTION:
 5 
5
1 i
2i (cos   i sin  )
6
6
As we discussed earlier we have to first find the modulus and argument of given complex No.
(a) (1 – i)5
Modulus:
|(1 – i)5| = |1 – i|5 = (2)5 = 42
Principal argument:
arg(1 – i)5 = 5 arg(1 – i)
tan  =
1
=1
1
y
x
O
  = – 4
1

y
x
1
P(1, 1)
arg(1 – i)5 = 5(–  4 ) =  5 4
Principal argument must lie in the domain (–, ], so add or
subtract multiples of 2 to bring  5 4 in above domain.
 Principal argument =  5 4 + 2 = 3 4
NOTE: In trigonometric and exponential form, we should write ‘’ as principal argument
Trigonometric form:
r(cos  + i sin )  42(cos 3 4 + i sin 3 4 )
Exponential form:
r·ei  42 e
i 3
4
(b) cos  3 – i sin  3
In such problems instead of first finding modulus and then finding argument, we can directly
adjust this and bring to trigonometric form.
We know, trigonometric form is r(cos  + i sin )
Now,
cos  3 – i sin  3
 cos(–  3 ) + i sin (–  3 )
 1·[cos(–  3 ) + i sin (–  3 )]
Since –  3 belongs to principal argument domain, the above is in trigonometric form.
Exponential form:
r·ei  1· e
i 
3
COMPLEX NUMBERS
[28]
(c) – cos  3 + i sin  3
We have to change sign of cos but not sin. Thus we should work in 2nd quadrant with 
– cos  3 + i sin  3
Now,
 cos( –  3 ) + i sin ( –  3 )
 1·(cos 2 3 + i sin 2 3 )
This is trigonometric form, with modulus = 1 and principal argument = 2 3
i 2
r·ei  1· e
Exponential form:
(d) 2(sin  3 + i cos  3 )
3
We have to change sin to cos and cos to sin.
So, we have to work in 1st quadrant and with  2
2(sin  3 + i cos  3 )
Now,
 2[sin (  2 –  3 ) + i cos (  2 –  3 )]
 2·(cos  6 + i sin  6 )
This is in standard trigonometric form, with modulus = 2 and principal argument =  6
i
r·ei  2· e 6
Exponential form:
(e) –2(cos  6 + i sin  6 )
This is not in trigonometric form because modulus cannot be negative. So we rearrange
–2(cos  6 + i sin  6 )  2(–cos  6 – i sin  6 )
We have to change sign of both cos and sin, so we have to work in 3rd quadrant with 
 2[cos( +  6 ) + i sin( +  6 )]
 2·(cos 7 6 + i sin 7 6 )
Here, modulus = 2 and argument = 7 6 . Hence Principal argument = 7 6 – 2 =  5 6
Trigonometric form:
2·[cos (  5 6 ) + i sin (  5 6 )]
Exponential form:
r·ei  2· e
i (5 )
6
(f) 2(cos 300o – i sin 300o)
Now rearranging,
2(cos 300o – i sin 300o)
 2[cos(2 – 300o) + i sin(2 – 300o)]
 2·(cos 60o + i sin 60o)
This is in standard trigonometric form, with modulus = 2 and principal argument = 60o = 3
Exponential form:
i
r·ei  2· e 3
COMPLEX NUMBERS
[29]
(g) 1  cos  8   i sin 8

 5 
{Important}
5
First let us simplify:

8
 8 
1  cos  5   i sin 5
 

 2cos2
 2cos
4
4
4
 i 2sin cos
5
5
5
4  4
4 
cos  i sin 

5 
5
5
This is in trigonometric form with modulus 2·cos( 4 5 ) and principal argument 4 5
This is wrong because modulus cannot be negative but cos( 4 5 ) is negative. So we have
to bring it to proper form
 2cos(4 5)  cos 4 5  i sin 4 5 


Positive
Now, we should change sign of cos and sin from negative to positive. So, we have to work
in 3rd quadrant with .
 2cos(4 5 ) cos(  4 5 )  i sin(  4 5 )
 2cos(4 5 ) cos 9 5  i sin 9 5  Trigonometric form


Positive
Modulus = –2·cos( 4 5 ) and principal argument = 9 5 – 2 =   5
Exponential form:
(h)
r·ei  –2·cos( 4 5 )· e
i 
5
1 i

2i (cos  i sin  )
6
6
Here, for 1 + i
r = 2 and  =  4
In such problems write all complex numbers in exponential form

1 i
2(i)(cos   i sin  )
6
6
i

2(e
Trigonometric form:
2 e
4
i 
i
2)e 6

1 i (  4   2  6 )
e
2

1 i (712)
e
This is exponetial form
2
1
(cos 7  i sin 7 )
12
12
2
{ –i = e
i 
2}
COMPLEX NUMBERS
[30]
Illustration 33:
Find the value of:
SOLUTION:
(cos   i sin  )4
6
6
(b)


(sin  i cos )3
3
3
(a) (1 + i)20 – (1 – i)20
Whenever we have problems involving powers and simplification, it is better to use exponential form
(a)
(1 + i)20 – (1 – i)20
For (1 + i)
:
y
x
 = 4
i
Exponential form is r·ei = 2 e 4
For (1 – i)
:
O
y
y
r = 1 1 = 2
 = – 4
x
Exponential form is r·ei = 2 e
Now,
P(1, 1)
1

x
1
r = 1 1 = 2
i 
O
4
(1 + i)20 – (1 – i)20
i
= (2 e 4 )20 – (2 e
1

y
i 
x
1
P(1, 1)
4 )20
= 210 ei5 – 210 e–i5
= 210(ei5 – e–i5)
= 210{(cos 5 + i sin 5) – (cos 5 – i sin 5)}
= 210(2i sin 5) = 0
(b)
(cos   i sin  )4
6
6


(sin  i cos )3
3
3
For cos  6 – i sin  6 = cos(–  6 ) + i sin(–  6 )  r = 1 and  = –  6
Exponential form is r·ei = 1 e
i 
6
For sin  3 – i cos  3 = cos(  2 –  3 ) + i sin(  2 –  3 )
= cos  6 + i sin  6  r = 1 and  =  6
i
Exponential form is r·ei = 1 e 6
Given quantity becomes
=e
(e
i 
i
4 )4
(e 6 )3
=
e
i 2 
i
e
3
2
i (2  )
i (  )
3 2 = e
6
= cos( +  6 ) – i sin( +  6 )
= – cos  6 + i sin  6
=  32  i2
{ e–i = cos  – i sin )
COMPLEX NUMBERS
[31]
Illustration 34:
Solve the following equation:
x2 – y2 – i(2x + y) = 2i
SOLUTION:
We know that two complex numbers are equal iff their real part and imaginary parts are separately
equal.
x2 – y2 – i(2x + y) = 0 + 2i
Now,
x2 – y2 = 0  x2 = y2 or x = ±y
...... (1)
and
–(2x + y) = 2  2x + y = –2
...... (2)
From (1) putting y = x in (2), we get
2x + x = –2
 x =  2 3 and y =  2 3
From (1) putting y = –x in (2), we get
2x – x = –2
 x = –2 and y = 2
Thus, solution set of (x, y) is ( 2 3 ,  2 3 ) and (–2, 2)

CUBE ROOTS OF UNITY:
In complex numbers, we will study later in intermediate that one of the main applications of complex numbers is
to find all the roots of any number. Presently, we will study only Cube Roots of Unity.
Cube roots of unity is nothing but 3 roots of the equation x3 = 1
Solving,
x3 = 1
 x3 – 1 = 0
 (x – 1)(x2 + x + 1) = 0
 one real root is x – 1 = 0
x=1
and other two roots are roots of the equation x2 + x + 1 = 0
 x
1  1  4
2
x=
1  3i 1  3i
,
2
2
Thus, cube roots of unity are 1,
These two are complex roots
1  3i 1  3i
,
(one real root and two complex roots)
2
2
COMPLEX NUMBERS
We call
[32]
1  3i
=  then squaring both sides we get,
2
2
 1  3i 
2

 =

2 

1  3i 2  2 3i
= 2
4
 2 =
2  2 3i
1  3i
=
4
2
 the two complex roots are square of one-other
Thus, we can write cube roots of unity as 1, , 2 where  =

1  3i
1  3i
and 2 =
2
2
PROPERTIES OF CUBE ROOTS OF UNITY:
1.
3 = 1. Because  is root of x3 = 1 it will satisfy the equation.
2.
3n = 1; 3n + 1 = (3n)· = ; 3n + 2 = (3n)· = 2
Thus, we can bring any higher power of ‘’to one of the lower one.
3.
1 +  + 2 = 0
Because  is root of x2 + x + 1 = 0, it will satisfy the equation,
1 +  = –2 (or) 1 + 2 = – (or)  + 2 = –1
Illustration 35:
Find the values of:
(a) 35
SOLUTION:
(b) 1 + 17 + 31
(c) (3 + 2 + 4)5
In problems relating to cube roots of unity, we should extensively use the property 3 = 1 and
1 +  + 2 = 0
(a)
35 = 33 × 2 = (3)11 × 2 = 1 · 2 = 2 =
(b)
1 + 17 + 31
1  3i
2
= 1 + (3)5·2 + (3)10·
(c)
= 1 + 2 + 
{ 3 = 1}
=0
{1 +  + 2 = 0}
(3 + 2 + 4)5
= (3 + 2 + 3·)5
= (3 + 2 + )5
= (3 – 1)5
= 25 = 32
{  + 2 = –1}
COMPLEX NUMBERS
[33]
Illustration 36:
Find the value of (1 +  – 2)3 – (1 –  + 2)3
SOLUTION:
We know, 1 +  = –2 and 1 + 2 = –
(1 +  – 2)3 – (1 –  + 2)3
= (–2 – 2)3 – (– – )3
= (–22)3 – (–2)3
= (–2)36 – (–2)33
= –8 – (–8)
{ 3 = 6 = 1}
=0
Illustration 37:
If 1, , 2 are the cube roots of unity prove that (2 – )(2 – 2)(2 – 10)(2 – 11) = 49
SOLUTION:
L.H.S. = (2 – )(2 – 2)(2 – 10)(2 – 11)
= (2 – )(2 – 2)(2 – )(2 – 2)
(  9 = 1)
= (2 – )2(2 – 2)2
= (4 + 2 – 4)(4 + 4 – 42)
= (4 + 2 – 4)(4 +  – 42)
= 16 + 4 – 162 + 42 + 3 – 44 – 16 – 42 + 163
= 33 – 16( + 2)
= 33 – 16(–1) = 49 = R.H.S.
Illustration 38:
If 1, , 2 are the cube roots of unity prove that
(1 –  + 2)6 + (1 – 2 + )6 = 128 = (1 –  + 2)7 + (1 +  – 2)7
SOLUTION:
L.H.S. = (1 –  + 2)6 + (1 – 2 + )6
= (–2)6 + (–22)6
= 26 6 + 26 12
= 26·1 + 26·1
= 64 + 64 = 128
R.H.S.= (1 –  + 2)7 + (1 +  – 2)7
= (–2)7 + (–22)7
= (–2)7(7 + 14)
= (–2)7( + 2) = –27(–1)
= 27 = 128
[1 +  + 2 = 0 and 3n= 1]
COMPLEX NUMBERS
[34]
Illustration 39:
If ,  are the complex cube roots of unity, find the value of (1 – )(1 – )(1 – 2)(1 – 4).
SOLUTION:
Let  =  and  = 2. Now substituting for  and  in the expression
(1 – )(1 – )(1 – 2)(1 – 4)
= (1 – )(1 – 2)(1 – 4)(1 – 8)
= (1 – )(1 – 2)(1 – )(1 – 2)
(4 = , 8 = 2)
= [(1 – )(1 – 2)]2 = (1 – 2 –  + 3)2
= [2 – ( + 2)]2
=9
Illustration 40:
If ,  are the complex cube roots of unity, show that
2
a  b  c a  b  c
b  c  a  c  a  b

 

SOLUTION:
Put  =  and  = 2.
 a  b  c

b  c  a
2
L.H.S.= 
 a3  b c2 
=
2 
 b  c a 
2
 (a2  b  c)
= 
2 
 b  c a 
[  a = a3]
2
= ()2 = 2
a  b  c

c  a  b
R.H.S.= 
 a3  b4  c2 
2 
 c  a b 
=
 2  a b2  c 
=  
2

  c  a b 
= 2
[  a = a3, b = b4]
COMPLEX NUMBERS
[35]
Illustration 41:
If 1, , 2 are the cube roots of unity prove that
(1 –  + 2)(1 – 2 + 4)(1 – 4 + 8)(1 – 8 + 16) = 16
SOLUTION:
Since 1 +  + 2 = 0 and 8 = 3·3·2 = 2
L.H.S. = (1 –  + 2)(1 – 2 + 4)(1 – 4 + 8)(1 – 8 + 16)
= (–2)(–22)(–2)(–22)
= 24 · 3 · 3
= 24
= 16
= R.H.S.
Illustration 42:
If p and q be the roots of equation y2 + y + 1 = 0, then find the value(s) of pn + qn for n  N
SOLUTION:
The roots of y2 + y + 1 = 0 is given by
1  3i
1  3i 1  3i
,

2
2
2
These are nothing but the two complex cube roots of unity.
So, we can write p =  and q = 2.
Now,
pn + qn = n + 2n
For n = 1,
1 + 2 = –1
For n = 2,
2 + 4 = 2 +  = –1
For n = 3,
3 + 6 = 1 + 1 = 2
For n = 4,
4 + 8 =  + 2 = –1
Thus, when n is a multiple of 3, the expression is equal to 2 otherwise it is equal to –1
COMPLEX NUMBERS
[36]
Illustration 43:
If 1, , 2 are the cube roots of unity then find the value of
SOLUTION:
1  2  32
2  3  2

2  3  2 3    22
We know 3 = 1. We will use this property.
1  2 32
2  3 2

2  3 2 3   22
=
1 3  2 32
2  3 2

2  3  3 2
3   22
(2  3 2 ) (22  3  )

=
2  3 2
3   22
=+
= 2
Illustration 44:
If 1, , 2 are the cube roots of unity then find the value of
SOLUTION:
1  2  32
2  3  2

2  3  2 1  2  32
We know 3 = 1. By using this property we get,
1  2 32
2  3 2

2  3 2 1  2 32
=
1 3  2 32
2  3 2

2  3  3 2
1  2 32
2
3
(2  3 2 )  (2   1)

=
2  3 2
3   22
3
=+
2 (2 3  
=  + 2
= –1

3   22
 1)
{ 3 = 1}
COMPLEX NUMBERS
[37]
Illustration 45:
Find the value of: 1·(2 – )(2 – 2) + 2·(3 – )(3 – 2) + .................. + 7·(8 – )(8 – 2)
SOLUTION:
Given 1·(2 – )(2 – 2) + 2·(3 – )(3 – 2) + .................. + 7·(8 – )(8 – 2)
The above can be written in summation form as —
8
 (r 1)(r  )(r  2 )
r 2
8
=  (r  1)(r 2  r  1)
r 2
8
=  r3  r 2  r  r 2  r 1
r 2
8
= r 1
3
r2
8
= (23 + 33 + 43 + ............. + 83) – 1
2
= –13 + (13 + 23 – 83) – 7
= –1 +
8  9  (17)
–7
6
= –8 + 204
= 196
COMPLEX NUMBERS

[38]
MISCELLANEOUS ILLUSTRATIONS:
Illustration 46:

3 

3 
If zr = cos  r   i sin  r  then find the value of z1·z2·............ ·z
SOLUTION:
We know the form cos  + i sin  = ei.

3 
i r
3

3 
So,
zr = cos  r   i sin  r  = e
For r = 1,
z1 = e 3
For r = 2,
z2 = e
i
i 2
3 and so on
i

i 2
i
3 · ........... · e 3
z1·z2·.........·z = e 3 · e
=
  

i   2  3 ............... 
3 3 3


e
  
i 3 
1 1 
3

{a =  3 , r = 13 }
=e
  
i 3 
2
 3 
=e
i
=e 2
= cos  2 + i sin  2
=i
Illustration 47:
If | z1 | = | z2 | = | z3 | = 1 and |z1 + z2 + z3| = 3 2 then find the value of
SOLUTION:
1 1 1
 
.
z1 z2 z3
We use the property, if | z | = 1 then | z |2 = 1 or z · z = 1  z =
1
z
Using property,
|z1| = 1  z1· z1 = 1 or
1
= z1
z1
Similarly,
|z2| = 1  z2· z2 = 1 or
1
1
= z2 and = z3
z2
z3
Substituting in
1 1 1
 
we get | z1  z2  z3 |
z1 z2 z3
= | z1  z2  z3 |
= |z1 + z2 + z3| = 3 2
{ | z |  | z | }
COMPLEX NUMBERS
[39]
Illustration 48:
If x = cos A + i sin A, y = cos B + i sin B, z = cos C + i sin C, where A, B, C are the angles of a triangle, find the
value of xyz.
SOLUTION:
We know that,
x = cos A + i sin A = eiA
y = cos B + i sin B = eiB
z = cos C + i sin C = eiC
Thus,
xyz = ei(A + B + C)
 xyz = ei
 xyz = cos  + i sin 
 xyz = –1 + i·0 = –1
Illustration 49:
If
3 z2
3z  5z2
is purely imaginary then find the value of 1
.
7 z1
3z1  5z2
SOLUTION:
Given,
3z2
= purely imaginary or
7 z1
3z2
z
= Ki  2 = Ki
7 z1
z1
Now, we want
 is some real constant
z
3 5 2
3z1  5z2
z1
=
z
3z1  5z2
35 2
z1
3  5i
= 3  5i
=
=
| 3  5i |
| 3  5i |
9  252
9  252
=1
COMPLEX NUMBERS
Assess
Yourself
[40]
01. Write the following in terms of ‘i’
(a)
121
4
(b) 4 9
(c)
36
5
02. Find the value of
(a) i357
(b) i2 + i4 + i6 +...........(2n + 1) terms
(c) i11 + i33 + i45
03. Write the following complex numbers in the form of A + iB
(a) (1+i) + (2 – 7i)
(b) (2 – i)2
(d) (1 – i) . i (3 + 5i)
(e)
(c) (3 + i)3
2  3i
1  5i
04. If (x + 2y) – i (2x – y) = 1 + i, find the values of x and y
05. If z = 2 + i, find the value of z2 + 3z – 9
06. Solve for x and y : (1 + i) x – (3 + i2) y = 2x + iy
07. If x + iy =
1
, find the value of 2x2.
1  cos   i sin 
3
3
1 i   1 i 
 
  a  ib , find the value of a + 2b
 1i   1 i 
08. If 
09. Write the conjugate of the following complex numbers
(a) 1+2i
(c)
3  4i
i
10. If z = 1 + 2i , find the values of zz and
(b) (1 + i) + (3 – 4i)
(d)
( 1  2i )( 3  5i )
1 i
z
z
11. Find the real and imaginary part of
(a)
(2  3i )(1 i )
2 i
(b)
12. Find the additive inverse of 2 + 5i.
13. Find the multiplicative inverse of 3 + 4i.
3
2  cos   i sin 
COMPLEX NUMBERS
[41]
14. Find the modulus of the following complex numbers
(a) 3 – 7i
(c)
(e)
1
2  5i
(b)
 3  4i  ( 2  i )
(d) (3 – i)4
( 1  2i )
 2  i  ( 1  i )6
2
(f)
3
( 2  i ) ( 4  3i )
1
1  cos   i sin 
15. Find x2 + y2 in each case
(2 i )
(a) (1 + i) 4 = x + iy
(b)  3  i  ( 4  2i )  x  iy
(c) (2 + i) (3 – 4i) = x – iy
(d) (1 – i) (1 – 2i) (1 – 3i) ..........(1 – ni) = x – iy
16. If (a1 + ib1) (a2 + ib2) +...............(an+ibn) = x + iy, find the value of
(a12 + b12) (a22 + b22) +..........(an2 + bn2)
17. Find the principal arguement and general argument for the following complex numbers
(a) 3 + i
(b) 3 – 3i
(c) –1 +3 i
(d) –5/2
(e) –i/5
(f) 1 – i
18. Find the modulus and principal argument of the following complex numbers
(a) – i)4
(b)
3 i
1 3i
zz 
 z3 
19. If /6, /4, –/3 are arguments of z1, z2 , z3 find the value of amplitude  1 2  .
20. Write the following in trigonometric and exponential form
(a) –3 i
(b)
1 i
3
21. Simplify :
(a)
(cos   i sin )2 (cos 2  i sin 2 )
(cos   i sin )3
(b) (1+ i)5
(c) ( 3  i )4
(c)
1 i
1  3i
COMPLEX NUMBERS
[42]
22. Write the following in trigonometric and exponential form
(a) 3 i)4
(b) – cos /6 + i sin /6
(c) – cos /3 – i sin /3
(d) –2 (sin /3 + i cos /3)
(e) 2 (cos 270° + i sin 270°)
(f) sin /4 – i cos /4
(g) 2 (1 +cos 6/5 +i sin 6/5) (h)
1  3i
i(cos   i sin  )
3
3
23. Find the value of
(a) 3 i)10 + (3 – i)10
(b)
(cos   i sin  )3
4
4


i(sin
 i cos
)2
3
3
24. Find the value of
(a) 23
(b) 1 + 23 +76
(c) (2 – –2)5
25. Find the value of
(a) – – – – 
(b) (1+ )(1+ 2) (1+ 4) (1+ 5)...................2n factor
(c) + –6+ – 
(d)
a  b  c 2
c  a  b 2
26. If ,  were complex cube roots of unity, then find the value of
(a) ––
 a  b  c 

 b  c  a 
3
(b) 
27. If z1  z2  z3  1 , such that z1  z2  z3  1 find the value of
28. If
2z  7z2
5z2
is purely imaginary, then find the value of 1
3z1
2z1  7z2
29. If ,  are real numbers find the value of
~ ~ ~
  i
  i
1 1 1
 
z1 z2 z3
COMPLEX NUMBERS
[43]
ANSWER KEY
ASSESS YOURSELF
01.
(a) 11 2 i
(b) 6i
(c) 6i 5
02.
(a) i
(b) – 1
(c) i
03.
(a) 3 – 6i
04.
x = –1/5 ; y = 3/5
05.
7i
06.
No solution
07.
1/2
08.
–4
09.
(a) 1 – 2i
(b) 4 – 3i
10.
(a) 5
(b) –3/5 + 4/5 i
11.
(a) –7/5 + 9/5 i
(b)
12.
– 2 – 5i
13.
1/(3 + 4i)
14.
(a) 58
(b) 1/29
(c) 5 (d) 100 (e) 85/625 (f) 2cos 
15.
(a) 16
(b) 1/40
(c) 125 (d) (1 + 12) (1 + 22) (1 + 32)..........(1 + n2)
16.
x2 + y2
17.
(a)  6 ; 2n  6
(b) 3 – 4i (c) 18 + 28i (d) –2 + 8i (e) 17/26 – 7/26 i
(c) 4 + 3i
(d) 7 + 6i
6  3cos 
3sin 
i
5  4cos  5  4cos 
1
(b)   4 ; 2n  4
,  -,
2
i
(c) 2[cos3isin 3]; 2e 3 (d) ; (2n  1)
(e)   2 ; 2n   2
18.
(a) 4, -(b) 1, /2
19.
–5/12
20.
(a) 2[cos   i sin  ]; 2e
3
3
i

3
(b)

(c)
1
1 i 12
[cos( )  i sin( )];
e
12
12
2
2
2
[cos( )  i sin( )];
4
4
3

2 i 4
e
3
COMPLEX NUMBERS
[44]
21.
(a) e i 7 or cos i sin 7 (b) – 4 – 4i
22.
(a) 16 [cos 2  i sin 2 ]; 16e
3
3
(c) cos 2  i sin 2 ; e
3
i
3
(e) 2[cos( )  i sin(
2


(g) 4cos 3 5  cos
i
2
(c) 8 (–1–3i)
2
3
i
6
2
3
6
4
2
24.
(a) (b) 0 (c) 35
25.
(a) (b) 1 (c) 128 (d) 2
26.
(a) 0
27.
1
28.
1
29.
1
4
i

(h) 2(cos   i sin  ); 2e 6
6
6
(b) 1
~ ~ ~
i
6
(f) [cos( )  i sin( )]; e
2
2 
3 i 5
 i sin  ; -4cos
e
5
5 
5
(a) 0 (b) e –i /12
6
(d) 2[cos 5  i sin 5 ]; 2e

i
)]; 2e 2
23.
5
(b) cos(5 )  i sin(5 ); e 6
i

4
5
6