Complex Analysis - Maths KU

196 Chapter 4 Elementary Functions
Note
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Complex powers defined by (1) satisfy the following properties that are
analogousto propertiesof real powers:
and
z α1 z α2 = z α1+α2 , zα1
z α2 = zα1−α2 ,
(z α ) n = z nα
for n =0, ±1, ±2,... .
Each of these properties can be derived from Definition 4.4 and Theorem 4.2.
For example, by Definition 4.4, we have z α1 z α2 = e α1 ln z e α2 ln z . Thiscan be
rewritten as z α1 z α2 = e α1 ln z+α2 ln z = e (α1+α2)lnz by using Theorem 4.2(ii).
Since e (α1+α2)lnz = z α1+α2 by (1), we have shown that z α1 z α2 = z α1+α2 .
Not all propertiesof real exponentshave analogouspropertiesfor complex
exponents. See the Remarksat the end of thissection for an example.
Principal Value of a Complex Power Aspointed out, the complex
power z α given in (1) is, in general, multiple-valued because it is defined
using the multiple-valued complex logarithm ln z. We can assign a unique
value to z α by using the principal value of the complex logarithm Ln z in
place of ln z. Thisparticular value of the complex power iscalled the principal
value of z α . For example, since Ln i = πi/2, the principal value of i 2i
isthe value of i 2i corresponding to n = 0 in part (a) of Example 1. That is,
the principal value of i 2i is e −π ≈ 0.0432. We summarize this discussion in
the following definition.
Definition 4.5 Principal Value of a Complex Power
If α isa complex number and z �= 0, then the function defined by:
z α = e αLnz
iscalled the principal value of the complex power z α .
The notation z α will be used to denote both the multiple-valued power
function F (z) =z α of (4) and the principal value power function given
by (6). In context it will be clear which of these two we are referring to.
EXAMPLE 2 Principal Value of a Complex Power
Find the principal value of each complex power: (a) (−3) i/π (b) (2i) 1−i
Solution In each part, we use (6) to find the principal value of z α .
(a) For z = −3, we have |z| = 3 and Arg(−3) = π, and so Ln (−3) = log e 3+iπ
by (14) in Section 4.1. Thus, by identifying z = −3 and α = i/π in (6),
we obtain:
or
(−3) i/π = e (i/π)Ln(−3) = e (i/π)(log e 3+iπ) ,
(−3) i/π = e −1+i(log e 3)/π .
(6)
(5)