Definition of $a^b$ for complex numbers Problem statement
Let $\Omega \subset C^*$ open and let $f:\Omega \to \mathbb C$ be a branch of logarithm, $b \in \mathbb C$, $a \in \Omega$. We define $a^b=e^{bf(a)}.$
$(i)$ Verify that if $b \in \mathbb Z$, $a^b$ doesn't depend on the election of $f$ and coincides with $a...a$ ($b$ times).
$(ii)$ Let $z \in \Omega, \space a,b \in \mathbb C$ such that $z^a \in \Omega$. What is the relation between $z^{a+b}$ and $z^az^b$? And what relation is there between $z^{ab}$ and $(z^a)^b$? And if $b \in \mathbb Z$?
The attempt at a solution
If $f$ is a branch of logarithm, then $f(z)=log(|z|)+iarg(z)+i2k\pi$ for $k$ a fixed integer.
For $(i)$, $$a^b=e^{b(log(|a|)+iarg(a)+i2k\pi)}$$ I don't know how to conclude that this expression equals to $a...a$ ($b$ times) just by the fact that $b \in \mathbb Z$. 
For $(ii)$, I could prove that $z^{a+b}=z^az^b$.
Now,$$z^{ab}=e^{ab(log(|z|+iarg(z)+i2k\pi)}$$
On the other hand, $${z^a}^b=e^{b(log|z^a|)+iarg(z^a)+i2k\pi)}$$
This last expression equals to $$e^{b(log|z^a|)+iarg(z^a)+i2k\pi)}=e^{b(alog|z|)+ia(arg(z))+i2k\pi)}$$
I don't know what to conclude from here: is it that $z^{ab}=(z^a)^b$ always or just for the case $b \in \mathbb Z$?
 A: For ii) Notice that , first of all, $f(z)=log|z|+ iargz+i2k\pi$ is well-defined within a branch , since the arguments over a fixed branch are all within , at most $2\pi$ from each other, so that, within a branch , $f(z)$ will have a unique value, because there will be exactly one value of $k$, so that the logz falls within that branch. This is why we choose a branch, so that the value logz is well-defined within that branch . 
Note that when you do $(z^a)^b$ , you multiply the argument by $b$, and that, when doing that, you may "jump a branch". Basically, $(z^a)^b= (z^a)^b \pm 2k\pi$ , to allow for the fact that the product $ab$ may be larger than the allowed argument within your branch, e.g., if you have an argument of $\pi/3$ in, say, $(- \pi, \pi]$, then, when raising to the power $b$, your argument will be multiplied by $b$, and may become larger than $\pi$. Then, in order to stay within your initial branch, you add/subtract copies of $2\pi$ (the period of the argument) , so that $ba$ is a value within your branch. So, say we have an argument $\pi/3$ in this branch, and we raise to the power of $9$. Then $9\pi/3=3\pi$ falls outside of the argument range, so, we subtract $2\pi$, for the argument to equal $\pi$, which is in $(-\pi, \pi]$.
A similar argument applies to $z^a z^b$; we have $z^az^b=z^a+b \pm 2k\pi$ , to allow for the fact that the sum of the arguments falls within the allowed branch argument; if the argument does not fall within this range, then we add multiples of $2k\pi$ until the argument does fall within the range of $argz$.
A: 
I don't know how to conclude that this expression equals to $a...a$ ($b$ times) just by the fact that $b\in\mathbb{Z}$.

If $\theta=Arg(a)$, then
$$\exp\{b\cdot(\ln(|a|)+(\theta+2k\pi)i)\},k,b\in\mathbb{Z}$$
$$=\exp\{b \ln(|a|)+b\theta i+2bk\pi i\},k,b\in\mathbb{Z}$$
$$=\exp\{b\ln(|a|)+b\theta i\}\cdot\exp\{2bk\pi i\}=(*)$$
$b,k\in\mathbb{Z}\Rightarrow \exp\{2bk\pi i\}=\exp\{2m\pi i\},m\in\mathbb{Z}\Rightarrow \exp\{2m\pi i\}=1\Rightarrow$
$$(*)=\exp\{b\cdot(\ln(|a|)+\theta i)\}\cdot 1$$
$$=\exp\{b\cdot f(z)\}=(**)$$
The last result is dependent only on the principal $\log$ branch $f(z)$, so in this case using your definition,
$$(**)=a^b=a\cdot a\cdot a\cdot\cdots a (b\text{-times})$$
