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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 $0<r<R$ 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 $A(0,r,R)\setminus \{z\in\mathbf{C}\mid r<\operatorname{Re} z<R,\operatorname{Im} z=0\}$.

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?

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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 $D=A(0,r,R)\setminus \{z\in\mathbf{C}\mid r<\operatorname{Re} z<R,\operatorname{Im} z=0\}$. 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 $$fe^{-(L+c)}=1.$$

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