(√
3 +j)−^3 =(
2(
cosπ
6+jsinπ
6))− 3
=1
8(
cosπ
6+jsinπ
6)− 3=1
8(
cos(
−π
2)
+jsin(
−π
2))
=−1
8jExercise on 12.7
By conversion to polar form and use of De Moivres’ theorem evaluate
(i) j^7 (ii) ( 1 +j)^5 (iii) (
√
3 −j)−^4Answer
(i) −j (ii) − 4 ( 1 +j) (iii) 2−^4
(
−1
2+√
3
2j)12.8 De Moivre’s theorem for fractional powers
We have considered simple algebra of complex numbers – addition, subtraction, multipli-
cation and ‘division’. So far, apart from
√
−1 we have not however considered taking the
roots of a complex number. We don’t have to look far to see that this requires some care.
A few examples will help:
x^2 =1 impliesx=±( 1 )1(^2) =±
√
1 =± 1
since(− 1 )^2 = 12 =1. Two values, as we might expect.
What aboutx^3 =1? In this case 1 is again a root, 1^3 =1, but−1won’tdobecause
(− 1 )^3 =− 1 =1. We would in any case expectthreevalues of the cube root, since we
expect three solutions to a cubic equation. In fact in this simple case we can actually find
them by some algebra.
x^3 − 1 =(x− 1 )(x^2 +x+ 1 )= 0
So
x= 1
or
x^2 +x+ 1 = 0
The latter equation gives twocomplexroots:
x=−
1
2
±j
√
3
2
So, in fact wedohave three cube roots of+1:
1 ,−
1
2
+j
√
3
2
,−
1
2
−j
√
3
2