170 CHAPTER 5 • ELEMENTARY FUNCTIONS
10. For what values of z is it true t hat(a) Log ( *) =Log (z1) - Log (z2)? Why?
(b} f;Log(z) =~?Why?
(c) Log ( ±) = -Log (z)? Why?- Construct branches off (z) = log (z + 2} that are analytic at all points in the plane
except at points on the following rays.
(a) {(x, y): x;::: -2, y = O}.
(b) {(x,y}: x= -2, y;::: O}.
(c) {(x,y): x = -2, y:;;; O}.- Show that the mapping w =Log (z) maps
(a) the ray { z = re'^8 : r > 0, () = f} one-tc:rone and onto t he horizontal
line { ( u, ti) : v = f}.
(b) the semicircle { z = 2e'^9 : -~ :;;; 8 :;;; ~} one-to-one and onto the verti-
cal line segment {(In 2, t1) : -~ :;;; t1:;;; ~}.- Find specific values of z1 and z2 so that Log(~):/= Log(z1)-Log(z2)-
1 4. Show why the solutions to Equation (5-10) are given by those in Equation (5-11).
Hint: Mimic the process used in obtaining Identities (5-8) and (5-9).
15. Explain why no branch of the logarithm is defined when z = 0.
5.3 Complex Exponents
In Section 1.5 we indicated that it is possible to make sense out of expressions
such as JI + i or i; without appealing to a number system beyond the framework
of complex numbers. We now show how this is done by taking note of some
rudimentary properties of the complex exponential and logarithm, and then using
our imagination.
We begin by generalizing Identity (5-15). Equations (5-12) and (5-14)show that log (z) can be expressed as the set log (z) = {Log (z) + i2mr: n is an
integer}. We can easily show (left as an ex~cise) that, for z =F 0, exp [log" (z)]
= z, where log"(z) is any branch of the function log(z). But this means that, for
any ( E log (z), the identity exp ( = z bolds true. Because exp [log ( z)] denotes
the set {exp(: ( E log(z)}, we see thatexp[Jog(z)] = z, for z=F O.
Next, note that Identity (5-17) gives log(zn) = nlog(z), where n is any
natural number, so that exp[log(zn)] = exp[nlog(z)] = zn, for z =F 0. With
these preliminaries out of the way, we can now come up with a definition of a
complex number raised to a complex power.