Page 124 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

114 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS

-1 +1

FIGURE 7.3: The roots of 1 1/2

In fact in this example θ is also π/2 + 2kπ. Using the fact that

z = re −i(θ+2kπ) k = 1, 2, 3,...

it is easy to show that

√ θ + 2πk θ + 2πk
z 1/n = n r cos + i sin (7.14)
n n
This is De Moivre’s theorem. For example when n = 2 there are two solutions and when
n = 3 there are three solutions. These solutions are called branches of z 1/n . A region in which
the function is single valued is indicated by forming a branch cut, which is a line stretching from
the origin outward such that the region between the positive real axis and the line contains
only one solution. In the above example, a branch cut might be a line from the origin out the
negative real axis.
Example 7.2. Find 1 1/2 and represent it on the polar diagram.

θ θ
1 1/2 = 1 cos + kπ + i sin + kπ
2 2
and since θ = 0 in this case

1 1/2 = cos kπ + i sin kπ

There are two distinct roots at z =+1for k = 0and −1for k = 1. The two values are
√
shown in Fig. 7.3. The two solutions are called branches of 1, and an appropriate branch cut
might be from the origin out the positive imaginary axis, leaving as the single solution 1.

Example 7.3. Find the roots of (1 + i) 1/4 .
√
π
Making use of Eq. (7.13) with m = 1and n = 4, r = 2,θ = 4 ,wefind that
√ π 2kπ π 2kπ
(1 + i) 1/4 = ( 2) 1/4 cos + + i sin + k = 0, 1, 2, 3
16 4 16 4