1550251515-Classical_Complex_Analysis__Gonzalez_

282 Chapter^5

lead to the impossibility 2 = 0.

To find the images of the horizontal segments z = x + im, -%7r < x <

(^1) /
2 7r, let ·tanhm =a, tanx = t, w = tanz = u +iv. '.1'hen from (5.19-14)
we get
(1 - a^2 )t
u = 1 + a2t2 '
which are the parametric equations of a circle with the point u = 0, v = 1/ a
missing. This point is approached as x --+^1 / 2 7r or t --+ oo (Fig. 5.25). By
eliminating the parameter t we get
u
2

• [v -1/2(a + ~ )r = 1/4 (a -~ r

which shows that the whole circle has a center at (0, %(a+ a-^1 )) and

radius r =%la -a-^1 1. For a= 0 the circle degenerates into the real axis

u = t, v = 0.

y

I {.)

I II

I x

I

t t

I

I

i Y = m

z Z1 Z2

---I ----- ---~-....... t-- -

-%'!1'1 I 0 %'!1' 'l1' 3'!1'/2 2'11' 5'11'/2 x

I

I I t

I

I

u

Fig. 5.25