# 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 $$0 and assume $$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.

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?

## 1 Answer

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