Choosing complex log branch 
Let $ D\subset \mathbb C\setminus\left\{ 0\right\}  $ be a simply connected region. Let $n\geq 2 $ be an integer. Let $ a\in D $ and let $ b \in \mathbb{C} $ such that $b^n=a$. Prove that there exists a unique holomorphic function $ f:D\to \mathbb{C} $ such that $ f(a)=b $ and $ (f(z))^n=z $ for all $ z\in D $.

What I want to do, is to define a branch of $\log $. But Im not sure which would be the specific branch that I should choose.
I cant really see why this $ f $ really has to be unique, for example, what if I'll choose the following 2 branches of log, which for any $ e^{x}=y $ defined by:
$$\begin{align*}
\log_{1}\left(y\right)&=x\\
\log_{2}\left(y\right)&=x+2\pi i\cdot n\\
\end{align*}$$
Then, I'll define
$$\begin{align*}
f_{1}\left(z\right)&=e^{\frac{1}{n}\log_{1}\left(z\right)}\\ f_{2}\left(z\right)&=e^{\frac{1}{n}\log_{2}\left(z\right)}\\ 
\end{align*}$$
Both $ f_1, f_2 $ are holomorphic, and we have
$$\begin{align*}
f_{1}\left(a\right)&=e^{\frac{1}{n}\log_{1}\left(a\right)}=e^{\frac{1}{n}\log_{1}\left(b^{n}\right)}=e^{\frac{1}{n}\log_{1}\left(e^{n\log_{1}\left(b\right)}\right)}=e^{\frac{1}{n}n\log_{1}\left(b\right)}=b\\ 
f_{2}\left(a\right)&=e^{\frac{1}{n}\log_{2}\left(a\right)}=e^{\frac{1}{n}\log_{2}\left(b^{n}\right)}=e^{\frac{1}{n}\log_{2}\left(e^{n\log_{2}\left(b\right)}\right)}=e^{\frac{1}{n}\left(n\log_{2}\left(b\right)+2\pi i\cdot n\right)}=b 
\end{align*}$$
I guess that if something is wrong it is probably the last step where I claim that $ b^{n}=e^{n\log\left(b^{n}\right)} $. Im not sure if it holds for any branch of log, or just for one specific branch.
I would really appreciate an explanation about when we can say that $ b^{n}=e^{n\log\left(b^{n}\right)} $ (for what branches, and how can we prove it), and also an explanation for how to choose the right branch of logarithm so I'll have uniqueness for the $ f $ that I tried to define.
Thanks in advance.
 A: It seems that you know that on each simply connected region $D$ not containing  $0$ there exists a branch $\log_1 : D \to \mathbb C$ of the logarithm. If $\log_2$ is another branch, then $\phi(z) = \log_2 z -  \log_1 z$ has the property
$$e^{\phi(z)} = e^{\log_2 z} / e^{\log_1 z} = z/z = 1$$
and we conclude that $\phi(z)$ must be constant with value $2k\pi i$ for some $k \in \mathbb Z$. Now define
$$f_i(z) = \frac{b}{e^{\frac{1}{n}\log_i a}}e^{\frac{1}{n}\log_i z} . \tag{1}$$
Then
$$f_i(a) = b \quad, \quad (f_i(z))^n = \frac{b^n}{a}z = z .$$
Using $\log_2 z =  \log_1 z + 2k\pi i$ we see that
$$f_2(z) = \frac{b}{e^{\frac{1}{n}\log_2 a}}e^{\frac{1}{n}\log_2 z} = \frac{b}{e^{\frac{1}{n}(\log_1 a + 2k\pi i)}}e^{\frac{1}{n}(\log_1 z +2k\pi i)} = \frac{b}{e^{\frac{1}{n}\log_1 a} e^{\frac{2k\pi i}{n}}}e^{\frac{1}{n}\log_1 z} e^{\frac{2k\pi i}{n}} \\ = \frac{b}{e^{\frac{1}{n}\log_1 a}}e^{\frac{1}{n}\log_1 z} = f_1(z) .$$
This shows that the choice of a logarithm branch in $(1)$ is irrelevant.
However, the independence from the choice of a logarithm branch  does not prove that there exists a unique holomorphic $f$ as in your question.
Anyway, we have constructed such $f$. Let $g$  be another holomorphic function with this property. Consider $\psi(z) = f(z)/g(z)$. Then $\psi(a) = 1$ and $(\psi(z))^n = 1$ for all $z \in D$. Therefore $\psi$ is a continuous function taking values in the discrete set $R_n$ of $n$-th roots of unity. Such a function  must be constant, hence $\psi(z) = 1$ for all $z \in D$.
Update:
The OP asked for the intuiton to use $(1)$ as the definition of $f(z)$. The "obvious" attempt is to define $f(z) = e^{\frac{1}{n}\log z}$, where $\log$ is any branch of the logarithm on $D$. Then $(f(z))^n = z$, but we cannot expect that $f(a) = b$. So try $f(z) = c e^{\frac{1}{n}\log z}$, where $c$ is some $n$-th root of unity. Again $(f(z))^n = z$. Now we can determine $c$ via $b = f(a) = c e^{\frac{1}{n}\log a}$ which gives $c =  \frac{b}{e^{\frac{1}{n}\log a}}$. Clearly $c^n = \frac{b^n}{a} = 1$, i.e. $c$ is in fact an $n$-th root of unity.
The OP also says that we can choose a specific branch $\log^*$ of the logarithm on $D$ such that $f(z) = e^{\frac{1}{n}\log^* z}$ satisfies $f(a) = b$. That is correct. Start with an arbitrary branch $\log$. Since our above $c$ is an $n$-th root of unity, we have $c = e^{\frac{2k\pi i}{n}}$ for some $k \in \{0,\dots,n-1\}$. Then $\log^* z = \log z + 2k\pi i$ is the desired branch. Of course also each $\log_n^* z = \log z + 2(n+k)\pi i$ with $n \in \mathbb Z$ will do.
