1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

180 CHAPTER 5 • ELEMENTARY FUNCTIONS

We demonstrate that oos (z 1 + z 2 } = oosz 1 cos z 2 - sin z 1 sin z 2 by making

use of Identities (5-32)-(5-35):

OOSZi COSZ2 = ~ [ei(z1+z.} + ei(z1-z2} + ei(z2- z1} + e - i(z1+ z2)] and

• sin z 1 sin z 2 = ~ [ ei(%1+z2) ei(z1- z2) ei{z2-zi) + e-i(%1 +z2)].

Adding these expressions gives

which is what we wanted.

A solution to the equation f ( z} = 0 is called a zero of the given function f.

As we now show, the zeros of the sine and cosine function are exactly where

you might expect them to be.

• We have sinz = 0 iff z = nn, where n is any integer, and cosz = 0 iff
  z = ( n + D n, where n is any integer.
  We show the result for cos z and leave the result for sin z as an exercise.
  When we use Identity (5-35), cosz = 0 iff

0 = cosxcoshy - isinxsinhy.

Equating the real and imaginary parts of this equation gives

0 = cos x cosh y and 0 = sin xsinh y.

The real-valued function cosh y is never zero, so the equation 0 = oos x oosh y

implies that 0 = oosx, from which we obtain x = (n + 4) n for any integer

n. Using the values for z = x + iy = ( n + 4) n + iy in 0 = sin x sinh y yields

0 =sin [ ( n + ~) n] sinhy = (-l)"sinhy,

which implies that y = 0, so the only zeros for cos z are t hose given by

z = (n + 4) 7r for any integer n.

What does the mapping w = sinz look like? We can get a graph of the

mapping w = sin z = sin ( x + iy) = sin x cosh y + i oos x sinh y by using paramet-

ric methods. Let's consider the vertical line segments in the z plane obtained

by successfully setting x = 7 + ~~ for k = 0, 1,... , 12, and for ea.ch x value

and letting y vary continuously, -3 ::; y ::; 3. In the exercises we ask you to