Complex Analysis - Maths KU

Note: All values of i 2i are real.
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4.2 Complex Powers 195
In general, (1) gives an infinite set of values because the complex logarithm
ln z ismultiple-valued. When n is an integer, however, the expression in (1)
issingle-valued (in agreement with fact that z n isa function when n isan
integer). To see that this is so, we use Theorem 4.2(ii) to obtain:
z n = e n ln z = e n[log e |z|+i arg(z)] = e n log e |z| e n arg(z)i . (2)
If θ = Arg(z), then arg(z) =θ +2kπ where k isan integer and so
e n arg(z)i = e n(θ+2kπ)i = e nθi e 2nkπi .
From Definition 4.1 we have that e 2nkπi = cos(2nkπ)+i sin (2nkπ). Because
n and k are integers, it follows that 2nkπ isan even multiple of π, and so
cos(2nkπ) = 1 and sin (2nkπ) = 0. Consequently, e 2nkπi = 1 and (2) can be
rewritten as:
z n = e n log e |z| e nArg(z)i , (3)
which issingle-valued.
Although the previous discussion shows that (1) can define a single-valued
function, you should bear in mind that, in general,
z α α ln z
= e
definesa multiple-valued function. We call the multiple-valued function given
by (4) a complex power function.
EXAMPLE 1 Complex Powers
Find the valuesof the given complex power: (a) i 2i (b) (1+i) i .
Solution In each part, the valuesof z α are found using (1).
(a) In part (a) of Example 3 in Section 4.1 we saw that:
(4n +1)π
ln i = i.
2
Thus, by identifying z = i and α =2i in (1) we obtain:
i 2i = e 2i ln i = e 2i[(4n+1)πi/2] = e −(4n+1)π
for n =0,±1, ±2, ... . The valuesof i 2i corresponding to, say, n = −1,
0, and 1 are 12391.6, 0.0432, and 1.507 × 10 −7 , respectively.
(b) In part (b) of Example 3 in Section 4.1 we saw that:
ln(1 + i) = 1
2 log e 2+
(8n +1)π
i
4
for n =0,±1, ±2, ... . Thus, by identifying z =1+i and α = i in (1)
we obtain:
or
for n =0,±1, ±2, ... .
(1 + i) i = e i ln(1+i) = e i[(log e 2)/2+(8n+1)πi/4] ,
(1 + i) i = e −(8n+1)π/4+i(log e 2)/2 ,
(4)