Page 33 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes Now if we let w = Log z and w = Log z , and notice that w  w as z  z , this becomes
0 0 0 0
Log z Log z w w


lim 0  lim 0
w

z 0 z z z 0 w w 0 e  e w 0
1 1
= 
e w 0 z 0
Thus, Log is differentiable at z , and its derivative is z 1 0 .
0

We are now ready to give meaning to z , where c is a complex number. We do the obvious and
c
define
z = e c log z .
c
There are many values of log z, and so there can be many values of z . As one might guess,
c
e cLog z is called the principal value of z .
c
Note that we are faced with two different definitions of z in case c is an integer. Let’s see, if we
c
have anything to unlearn. Suppose c is simply an integer, c = n. Then
z = e n log z = e n(Log z + 2ki)
n
e
= e nLog z 2kni = e nLog z
There is, thus, just one value of z , and it is exactly what it should be: e nLog z = |z| e . It is easy
n
n in arg z
to verify that in case c is a rational number, z is also exactly what it should be.
c
Far more serious is the fact that we are faced with conflicting definitions of z in case z = e. In the
c
above discussion, we have assumed that e stands for exp(z). Now we have a definition for e that
z
z
implies that e can have many values. For instance, if someone runs at you in the night and hands
z
you a note with e written on it, how do you know whether this means exp(1/2) or the two
1/2
values e and  e ? Strictly speaking, you do not know. This ambiguity could be avoided, of
course, by always using the notation exp(z) for e e , but almost everybody in the world uses e z
x iy
with the understanding that this is exp(z), or equivalently, the principal value of e . This will be
z
our practice.
3.4 Summary
Let the so-called exponential function exp be defined by

exp(z) = ex(cos y + i sin y),
where, as usual, z = x + iy. From the Cauchy-Riemann equations, we see at once that this
function has a derivative every where—it is an entire function. Moreover,

d exp(z).
dz exp(z) 

Note next that if z = x + iy and w = u + iv, then
exp(z + w) = e [cos(y + v) + i sin(y + v)]
x+u
= e e [cos y cos v – sin y sin v + i(sin y cos v + cos y sin v)]
x u
= e e (cos y + i sin y) (cos v + i sin v)
x u
= exp(z) exp(w).

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