226 Higher Engineering Mathematics

Now try the following exercise

Exercise 90 Further problems on powers

of complex numbers

1. Determine in polar form (a) [1. 5 ∠ 15 ◦]^5
   (b)( 1 +j 2 )^6.
   [(a) 7. 594 ∠ 75 ◦(b) 125∠ 20. 61 ◦]


2. Determine in polar and cartesian forms
   (a) [3∠ 41 ◦]^4 (b)(− 2 −j)^5.
   [
   (a) 81∠ 164 ◦,− 77. 86 +j 22. 33
   (b) 55. 90 ∠− 47. 18 ◦, 38 −j 41

]

3. Convert( 3 −j)into polar form and hence
   evaluate( 3 −j)^7 , giving the answer in polar
   form. [

√

10 ∠− 18. 43 ◦, 3162∠− 129 ◦]

In problems 4 to 7, express in both polar and

rectangular forms.

4. ( 6 +j 5 )^3 [476. 4 ∠ 119. 42 ◦,− 234 +j415]


5. ( 3 −j 8 )^5
   [45530∠ 12. 78 ◦, 44400+j10070]


6. (− 2 +j 7 )^4 [2809∠ 63. 78 ◦, 1241+j2520]


7. (− 16 −j 9 )^6

[

( 38. 27 × 106 )∠ 176. 15 ◦,

106 (− 38. 18 +j 2. 570 )

]

21.3 Roots of complex numbers

Thesquare rootof a complex number is determined by

lettingn= 1 /2 in De Moivre’s theorem,

i.e.

√

[r∠θ]=[r∠θ]

1

(^2) =r
1
(^2) ∠
1
2
θ=
√
r∠
θ
2
There are two square roots of a real number, equal in
size but opposite in sign.
Problem 3. Determine the two square roots of the
complex number( 5 +j 12 )in polar and cartesian
forms and show the roots on an Argand diagram.
( 5 +j 12 )=
√
[5^2 + 122 ]∠tan−^1
(
12
5
)
= 13 ∠ 67. 38 ◦
When determining square roots two solutions result.
To obtain the second solution one way is to
express 13∠ 67. 38 ◦also as 13∠( 67. 38 ◦+ 360 ◦),i.e.
13 ∠ 427. 38 ◦. When the angle is divided by 2 an angle
less than 360◦is obtained.
Hence
√
( 5 +j 12 )=
√
[13∠ 67. 38 ◦]and
√
[13∠ 427. 38 ◦]
=[13∠ 67. 38 ◦]
1
(^2) and [13∠ 427. 38 ◦]
1
2
= 13
1
(^2) ∠
(
1
2
× 67. 38 ◦
)
and
13
1
(^2) ∠
(
1
2
× 427. 38 ◦
)

√
13 ∠ 33. 69 ◦and
√
13 ∠ 213. 69 ◦
= 3. 61 ∠ 33. 69 ◦and 3. 61 ∠ 213. 69 ◦
Thus, in polar form, the two roots are
3.61∠33.69◦and 3.61∠−146.31◦.
√
13 ∠ 33. 69 ◦=
√
13 (cos33. 69 ◦+jsin33. 69 ◦)
= 3. 0 +j 2. 0
√
13 ∠ 213. 69 ◦=
√
13 (cos213. 69 ◦+jsin213. 69 ◦)
=− 3. 0 −j 2. 0
Thus, in cartesian form the two roots are
±( 3. 0 +j 2. 0 ).
From the Argand diagram shown in Fig. 21.1 the two
roots are seen to be 180◦apart, which is always true
when finding square roots of complex numbers.
j 2
2 j 2
3
3.61
3.61
Imaginary axis
213.69 (^8) 33. 69 8
23 Real axis
Figure 21.1
In general,when finding thenthroot of a complex
number, there arensolutions. For example, there are
three solutions to a cube root, five solutions to a fifth
root, and so on. In the solutionsto the roots of a complex
number, the modulus,r,isalwaysthesame,butthe