Prove some properties of complex numbers raised to complex powers

If $z$ is a nonzero complex numbers and $w \in \mathbb{C}$ let

$z^w = e^{w \operatorname{Log} z},$

where

$\operatorname{Log} z = \log |z| + i \arg(z).$

Compute $1^i$ , $i^i$ , and $(-1)^i$ .
Prove that $z^a z^b = z^{a+b}$ if $a,b$ , and $z$ are in $\mathbb{C}$ with $z \neq 0$ .
What conditions on $z_1$ and $z_2$ must we have for the equation
$(z_1 z_2)^w = z_1^w z_2^w$

to hold? Show that the equation fails when $z_1 = z_2 = -1$ and $w = i$ .

The computations are as follows,
$\begin{align*} 1^i &= e^{i \operatorname{Log} 1} = e^0 = 1. \\[9pt] i^i &= e^{i \operatorname{Log} i} = e^{i \left( \frac{\pi i}{2} \right)} = e^{-\frac{\pi}{2}}. \\[9pt] (-1)^i &= e^{i \operatorname{Log} (-1)} = e^{i (\pi i)} = e^{- \pi}. \end{align*}$
Proof. Using the definitions we compute,
$\begin{align*} z^a z^b &= \left( e^{a \operatorname{Log} z} \right) \left( e^{b \operatorname{Log} z} \right) \\[9pt] &= e^{a \operatorname{Log} z + b \operatorname{Log} z} \\[9pt] &= e^{(a+b)\operatorname{Log} z} \\[9pt] &= z^{a+b}. \qquad \blacksquare \end{align*}$
First, if $z_1 = z_2 = -1$ and $w = i$ then we have
$(z_1 z_2)^w = 1^i = 1,$

but

$z_1^w z_2^w = (-1)^i \cdot (-1)^i = e^{-\pi} \cdot e^{-\pi} = e^{-2\pi}.$

Thus,

$(z_1 z_2)^w \neq z_1^w z_2^w.$

In order for $(z_1 z_2)^w = z_1^w z_2^w$ we must have $- \pi < \arg z_1 + \arg z_2 \leq \pi$ since

$(z_1 z_2)^w = e^{w \operatorname{Log} (z_1 z_2)} = z_1^w z_2^w \cdot e^{2n \pi i},$

and $n = 0$ only when $- \pi < arg z_1 + arg z_2 \leq \pi$ .

Stumbling Robot