262 COMPLEX NUMBERS- 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 ◦ 10 ′,38−j 41
]- Convert (3−j) into polar form and hence
evaluate (3−j)^7 , giving the answer in polar
form. [
√
10 ∠− 18 ◦ 26 ′, 3162∠− 129 ◦ 2 ′]In problems 4 to 7, express in both polar and
rectangular forms.- (6+j5)^3 [476. 4 ∠ 119 ◦ 25 ′,− 234 +j415]
- (3−j8)^5
[45530∠ 12 ◦ 47 ′, 44400+j10070] - (− 2 +j7)^4 [2809∠ 63 ◦ 47 ′, 1241+j2520]
- (− 16 −j9)^6
[
(38. 27 × 106 )∠ 176 ◦ 9 ′,106 (− 38. 18 +j 2 .570)]24.3 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+j12) in polar and carte-
sian forms and show the roots on an Argand
diagram.
(5+j12)=
√
[5^2 + 122 ]∠arctan
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+j12)=
√
[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 ◦ 41 ′and 3. 61 ∠ 213 ◦ 41 ′
Thus, in polar form, the two roots are
3.61∠ 33 ◦ 41 ′and 3.61∠− 146 ◦ 19 ′.
√
13 ∠ 33. 69 ◦=
√
13( cos 33. 69 ◦+jsin 33. 69 ◦)
= 3. 0 +j 2. 0
√
13 ∠ 213. 69 ◦=
√
13( cos 213. 69 ◦+jsin 213. 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. 24.1 the
two roots are seen to be 180◦apart, which is always
true when finding square roots of complex numbers.
Imaginary axis
j 2
213 ° 41 ' 33 ° 41 '
3.61
3.61
−j 2
− 3 3 Real axis
Figure 24.1
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 solutions to the roots of a
complex number, the modulus,r, is always the same,