mapping properties of $(1−z)^i$ There is a question on the complex analysis of Conway "Functions of one complex variable" (pg.$57\ \#28$) about the mapping properties of $(1−z)^i$. Can anyone describe me what does the exponent '$i$' mean? What does it do to $(1−z)$? I am confused. Can i write in the following way?
$$(1−z)^i=e^{i\cdot\log(1−z)}=\cos(\log(1−z))+i\sin(\log(1−z))$$
 A: As you mentioned, $(1-z)^i = \exp(i \log(1-z))$. However, $\log$ is a multi-valued function, and different branches of $\log$ will give you different branches of $(1-z)^i$.  I'll assume we're using the principal branch of the logarithm, i.e. the one with imaginary part in $(-\pi, \pi]$. 
For any complex $z$,  $\log(1-z)$ is thus in the horizontal strip with imaginary part in $(-\pi, \pi]$.  Multiply it by $i$, and you get the vertical strip with real part in $(-\pi, \pi]$.  The exponential function maps this into the annulus 
$\{w: e^{-\pi} < |w| \le e^\pi\}$.  However, this mapping is not one-to-one, 
as $\exp$ is periodic with period $2 \pi i$.  $\exp$ will map
 each of the rectangles $R_k = \{x+iy: -\pi < x \le \pi, 2k\pi < y \le 2(k+1)\pi\}$
one-to-one onto the annulus.  Now $ i \log(1-z) \in R_k$ iff $\text{Re}(\log(1-z)) \in (2k\pi, 2(k+1)\pi]$, and since $\text{Re}(\log(1-z)) = \log |1-z|$
this says $2k\pi < \log |1-z| \le 2(k+1)\pi$, or $e^{2k\pi} < |1-z| \le e^{2(k+1)\pi}$.
Thus one answer is that $(1-z)^i$ maps each annulus $e^{2k\pi} < |1-z| \le e^{2(k+1)\pi}$ one-to-one onto the annulus $e^{-\pi} < |w| \le e^{\pi}$.
