$f$ nowhere zero and holomorphic in the annulus $1<|z|<2$, and for each radius $r$ between 1 and 2, $\arg f$ is constant on $|z|=r$. Show $f$ constant
Suppose $f$ is a nowhere vanishing analytic function in the annulus $1<|z|<2$, and for each radius $r$ between $1$ and $2$, the argument of $f(z)$ is constant on the circle $|z|=r$. Show that $f$ is a constant function.
If $f$ were analytic on the whole disk, we know that the average of values on each radius is the same, so $f$ must map to a line and the open mapping theorem shows that $f$ must be constant, but I'm assuming this goes out the window because $z=0$ is not in the domain.
I suppose it would have to do with a locally defined logarithm $\log(f)$, by the assumption that $f$ is nowhere vanishing, but I don't see how $\log(f)$ mapping to a horizontal line on $|z|=r$, lends to $f$ being constant. Any help is appreciated. Thank you.
complex-analysis
Daniel Fischer
asked Aug 6 '15 at 4:17
BlakeBlake
For each $r\in (1,2)$, let $C_r = \{ z : \lvert z\rvert = r\}$. Since $f$ has constant argument on $C_r$, we can write
$$f(re^{i\varphi}) = a_r(\varphi)\cdot e^{i\theta(r)}$$
with $a_r(\varphi) > 0$ for all $r,\varphi$. By Cauchy's integral theorem,
$$I(r) = \frac{1}{2\pi i} \int_{\lvert z\rvert = r} \frac{f(z)}{z}\,dz$$
is independent of $r$. But on the other hand, parametrising the circle by $re^{i\varphi}$ we obtain
$$I(r) = \frac{1}{2\pi} \int_0^{2\pi} a_r(\varphi) \,d\varphi\cdot e^{i\theta(r)},$$
and since $\int_0^{2\pi} a_r(\varphi)\,d\varphi > 0$, it follows that
$$\arg I(r) = \theta(r).$$
Since $I(r) \equiv \operatorname{const}$, it follows that $\theta(r) \equiv \operatorname{const}$, i.e. the argument of $f$ is constant on the whole annulus. By the open mapping theorem, it follows that $f$ is constant.
Daniel FischerDaniel Fischer
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