Page 123 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

COMPLEX VARIABLES AND THE LAPLACE INVERSION INTEGRAL 113
Noting that

2
2
2
2
z = r (cos θ − sin θ + i2sin θ cos θ)
1 1
= r 2 (1 + cos 2θ) − (1 − cos 2θ) + i sin 2θ
2 2
2
= r [cos 2θ + i sin 2θ]
We deduce that

z 1/2 = r 1/2 (cos θ/2 + i sin θ/2) (7.12)

In fact in general

z m/n = r m/n [cos(mθ/n) + i sin(mθ/n)] (7.13)

Example 7.1. Find i 1/2 .
Noting that when z = I, r = 1and θ = π/2, with m = 1and n = 2.
Thus

1
i 1/2 = 1 1/2 [cos(π/4) + i sin(π/4)] = √ (1 + i)
2

Note, however, that if

π π

w = cos + π + i sin + π
4 4
2
1
then w = i. Hence √ (−1 − i) is also a solution. The roots are shown in Fig. 7.2.
2

i

FIGURE 7.2: Roots of i 1/2