1550251515-Classical_Complex_Analysis__Gonzalez_

Elementary Functions 283

Next, to determine the images of the vertical lines z = c + iy, 0 < !cl <

%7r, -oo < y < +oo, let tanc = b, tanhy = t. Then from (5.19-14) we get

Cb: u = b(l - t2) v = (1 + b2)t -1 < t < 1

1 + b^2 t^2 ' 1 + b^2 t2 '

which are the parametric equations of an arc of a circle. The arc intersects

the u-axis at (b, 0) (fort= 0), approaches w = -i as t--+ -1, and w = i

as t --+ 1 (Fig. 5.25). Elimination of the parameter t gives

so the arc has center at (1/ 2 (b - b-^1 ), 0) and radius r =^1 / 2 lb + b-^1 1. For
b = 0 the arc degenerates into u = O, v = t ( -1 < t < 1 ), i.e., into the

interval (-i, i) along the imaginary axis.

It is clear from Fig. 5.25 that the arcs Ca and Cb do intersect at a point

w = tan(c+im) since either i or -i is interior to Ca. Conversely, any w not

on (-ioo, -i] U (i, ioo) belongs to the intersection of certain arcs Ca and Cb.

Returning to equation (5.19-16), which can be written as

tan [ ~ (2n + 1 )7r + iy] = i ( 1 + e 2 Y

2

_ 1 ) = -i ( 1 + 1 ~

2

: 2 Y )

we see that the image of each half-line z = %(2n + l)7r + iy, y > 0, is the

open interval ( i, ioo) along the v-axis, described from ioo to i as y increases
from 0 to +oo, while the image of the other half-line z =^1 / 2 (2n + 1 )7r + iy,
y < O, is the open interval (-ioo, -i), described from -i to -ioo as y

increases from -oo to 0 (in both cases endpoints excluded).

The preceding discussion shows that each w E C~ - { i, -i} is the image
of just one point z E So = {z: -%7r < Rez $ %7r, -oo < Imz < +oo}.
However, by the periodicity of tanz, any such w is also the image of the
points Zn = z + n7r.

Exercises 5 .4

1. Evaluate:
   (a) e2i; (b) e-l+i'1r/4; ( c) el-i'lr/6.


2. Find the real and imaginary components of w = eez.


3. Find the image of the strip

S={z: z=x+iy,x~0,0$y$7r}

under the mapping defined by w = e=.