Holomorphic branch of $\log f$ and $f^{1/n}$ I am doing old complex analysis questions to prepare me for the final exam. I came upon this question, from which I could not solve the last part.
Let $0&lt;r&lt;R$ and assume $f$ has no zeros in a neighbourhood of $A(0,r,R)$.
Let $\rho\in(r,R)$. Prove that there is a holomorphic branch of $\log f$ on $A(0,r,R)\setminus \{z\in\mathbf{C}\mid r&lt;\operatorname{Re} z&lt;R,\operatorname{Im} z=0\}$.
This I proved with a theorem in my book that says that if $D$ is a simply connected domain in $\mathbf{C}$ and $f$ has no zeros on $G$, then there exists a holomorphic branch of $\log f$ on $D$.

(a) Take such a holomorphic branch $\log f$. Calculate $\lim_{y\to 0^+}\left[\log f(\rho+iy)-\log f(\rho-iy)\right]$.
(b) Prove that there exist holomorphic branches of $f^{1/n}$ on $A(0,r,R)$.

I have no idea how to do (a). For (b) I thought of doing something like $g(z)=e^{\frac1n\log(f(z))}$. But then the theorem does not work anymore, since the annulus is not simply connected.
Can someone provide some help?
First, (b) is false;  $r=1$, $R=2$, $f(z)=z$ is a counterexample.

It's not clear exactly what counts as an answer to (a). Say $g=\log f$ in $D=A(0,r,R)\setminus \{z\in\mathbf{C}\mid r&lt;\operatorname{Re} z&lt;R,\operatorname{Im} z=0\}$. Then $g'=f'/f$. So, writing $g(p)-g(q)$ as the integral of $g'$ over a path from $q$ to $p$, it follows that $$\lim_{y\to0}(g(\rho+iy)-g(\rho-iy))=-\int_{|z|=\rho}\frac{f'(z)}{f(z)}\,dz,$$which may well be the expected answer.

Here's a generalization of a correct version:


  
    Suppose $V$ is an open set and $f\in H(V)$ has no zero. The following are equivalent:
    
    (a') There exists $\log f\in H(V)$
    
    (b') For every $n\in\Bbb N$ there exists $f^{1/n}\in H(V)$.
  


Of course it's trivial that (a') implies (b'). Assume (b').

Let $g_n=f^{1/n}$. Say $\gamma$ is a closed curve in $V$, and define $$I=\frac1{2\pi i}\int_\gamma\frac{f'}f,$$
$$I_n=\frac1{2\pi i}\int_\gamma\frac{g_n'}{g_n}.$$

Since $f'/f=ng_n'/g_n$ it follows that $$I=nI_n\quad(n\in\Bbb N).$$

Since $I$ and $I_n$ are integers this implies that $$I=0.$$
Since $I=0$ for every closed path  $\gamma$ there exists $L\in H(V)$ with $$L'
=f'/f.$$Hence $fe^{-L}$ is constant (wlog $V$ is connected), and so there exists $c$ with $$fe^{-(L+c)}=1.$$
First, (b) is false;  $r=1$, $R=2$, $f(z)=z$ is a counterexample.

It's not clear exactly what counts as an answer to (a). Say $g=\log f$ in $D=A(0,r,R)\setminus \{z\in\mathbf{C}\mid r&lt;\operatorname{Re} z&lt;R,\operatorname{Im} z=0\}$. Then $g'=f'/f$. So, writing $g(p)-g(q)$ as the integral of $g'$ over a path from $q$ to $p$, it follows that $$\lim_{y\to0}(g(\rho+iy)-g(\rho-iy))=-\int_{|z|=\rho}\frac{f'(z)}{f(z)}\,dz,$$which may well be the expected answer.

Here's a generalization of a correct version:


  
    Suppose $V$ is an open set and $f\in H(V)$ has no zero. The following are equivalent:
    
    (a') There exists $\log f\in H(V)$
    
    (b') For every $n\in\Bbb N$ there exists $f^{1/n}\in H(V)$.
  


Of course it's trivial that (a') implies (b'). Assume (b').

Let $g_n=f^{1/n}$. Say $\gamma$ is a closed curve in $V$, and define $$I=\frac1{2\pi i}\int_\gamma\frac{f'}f,$$
$$I_n=\frac1{2\pi i}\int_\gamma\frac{g_n'}{g_n}.$$

Since $f'/f=ng_n'/g_n$ it follows that $$I=nI_n\quad(n\in\Bbb N).$$

Since $I$ and $I_n$ are integers this implies that $$I=0.$$
Since $I=0$ for every closed path  $\gamma$ there exists $L\in H(V)$ with $$L'
=f'/f.$$Hence $fe^{-L}$ is constant (wlog $V$ is connected), and so there exists $c$ with $$fe^{-(L+c)}=1.$$