1550251515-Classical_Complex_Analysis__Gonzalez_

(jair2018) #1
Elementary Functions 263

inverse image, namely, z = a and z = oo, respectively. It is said, in the case

of those exceptional points, that the n inverse images coalesce into one.
The n inverse images of w # 0, oo are given by

Zk =a+* y'W =a+ ~ei(Argw+^2 k7r)/n (5.15-1)


fork= 0, 1 ... , n -1, by virtue of (1.12-7). For n > 2, the points Zk lie
at the vertices of a regular n-gon with center at the point z = a, while for
n = 2 the two points z 0 , z 1 are symmetric with respect to a. Whatever
the case, lzk -al = ~for all k.

Looking now into the mapping z----> was given by w = (z - a)n, n > 1,

we have

and argw = nArg(z - a) (5.15-2)

which shows that each circle lz - al= r is mapped into the circle lwl = rn


(Fig. 5.20). However, as the point z describes the circle lz - al = r once in
the positive direction, its image w moves around the circle lwl = rn n times,
also in the positive direction, since ari increment of 271" of some particular
value of arg(z - a) results in an increment of 27rn of the corresponding
value of argw, according to the second equation of (5.15-2). In addition,
equations (5.15-2) show that as z describes the ray

Lo= {z: 0 < lz - al < oo, Arg(z - a)= Oo}

its image w moves along the ray


Lri = {w: 0 < lwl < oo,argw = n8 0 }
Now consider the region

R = {z: 0 < lz - al< oo, 80 < Arg(z - a)< Oi}


y v

0 x rn u

Fig. 5.20

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