I'm trying to do this old qualifying exam problem from UW Madison.
Let $D^\ast=\{z\in\mathbb{C},0<|z|<1\}$ and $f$ be a non constant holomorphic function on $D^\ast$. Assume that $\text{Im} f(z)\geq 0$ if $\text{Im} z\geq 0$ and $\text{Im} f(z)\leq 0$ if $\text{Im} z\leq 0$. Prove that if $z\in D^\ast$ is not real, then $f(z)$ is not real. Show that if $z\in (-1,0)\cup(0,1)$, then $f'(z)\not=0$. Prove that $0$ is either a removable singularity with $f'(0)\not=0$ or $0$ is a simple pole of $f$.
What I have thought of so far:
If $z\in D^\ast$ and $\text{Im} z>0$, but $\text{Im} f(z)=0$, then apply the maximum modulus principle on $\{z|z\in D^\ast,\text{Im} z>0\}$ to $e^{if}$ to obtain a contradiction. This shows that if $z\in D^\ast$ and $\text{Im} z>0$, then $\text{Im} f(z)>0$. Similarly if $z\in D^\ast$ and $\text{Im} z<0$. Furthermore, the reflection principle shows that $f(\overline{z})=\overline{f(z)}$. If $z\in(-1,0)\cup(1,0)$, to show that $f'(z)\not=0$, we use the following fact:
fact: If $f$ is a complex valued function continuous on $\overline{D(0,R)}$ and holomorphic on $D(0,R)$, then for any $z\in D(0,R)$, we have $$f(z)=\int_0^{2\pi}i \text{Im} f(\xi)\frac{\xi+z}{\xi-z}\frac{d\theta}{2\pi}+K$$ for some constant $K$, where $\xi=Re^{i\theta}$. We can differentiate the expression to get an expression of $f'(z)$ in terms of $\text{Im} f$. This same fact shows that $f'(0)\not=0$ if $0$ is a removable singularity. If $0$ is a pole, $\text{Im} z$ is dominated by the imaginary part of $\frac{C}{z^n}$ for some $n\in\mathbb{Z}^+$ and $C\in\mathbb{R}$ when $|z|>0$ is small, then unless $n\not=1$, we can find some $z\in D^\ast$, $\text{Im} z>0$ such that $\text{Im} f(z)<0$. So if $0$ is a pole, then it is a simple pole.
My question is: how to show that $0$ is not an essential singularity? Thanks!!
Best Answer
Since $f$ is holomorphic on the punctured unit disk $\mathbb{D}^*$, it admits the following Laurent expansion
$$ f(z) = \sum_{n\in\mathbb{Z}} a_n z^n = g(z) + a_0 + h(z), $$
where $a_n \in \mathbb{R}$ for all $n \in \mathbb{Z}$ and
$$ g(z) = \sum_{n > 0} a_n z^n \qquad \text{and} \qquad h(z) = \sum_{n < 0} a_n z^n. $$
Note that $g(z)$ converges on all of the unit disk $\mathbb{D}$, whereas $h(1/z)$ defines an entire function. Now assume that $0 < r < 1$ and $r < |z| < 1$. By using the relation $\operatorname{Im}\{g(z)\} = -\operatorname{Im}\{g(\overline{z})\}$,
\begin{align*} &\int_{|\xi|=r} i \operatorname{Im}\{f(\xi)\} \biggl( \frac{z+\xi}{z-\xi} \biggr) \, \frac{|\mathrm{d}\xi|}{2\pi r} \\ &= \int_{|\xi|=r} i \operatorname{Im}\{h(\xi) - g(r^2/\xi)\} \biggl( \frac{z+\xi}{z-\xi} \biggr) \, \frac{|\mathrm{d}\xi|}{2\pi r} \\ &= \int_{|\zeta|=r} i \operatorname{Im}\{h(r^2/\zeta) - g(\zeta)\} \biggl( \frac{\zeta+r^2/z}{\zeta-r^2/z} \biggr) \, \frac{|\mathrm{d}\zeta|}{2\pi r} \tag{$\zeta=r^2/\xi$} \\ &= h(z) - g(r^2/z). \end{align*}
Here, the last step is a consequence of the Schwarz integral formula (of OP's version) applied to the holomorphic function $h(r^2/z) - g(z)$ on $\mathbb{D}$ and $|r^2/z| < r$. Then by substituting $\xi = re^{i\theta}$,
\begin{align*} h(z) - g(r^2/z) &= \int_{-\pi}^{\pi} i \operatorname{Im}\{f(re^{i\theta})\} \biggl( \frac{z+re^{i\theta}}{z-re^{i\theta}} \biggr) \, \frac{\mathrm{d}\theta}{2\pi} \\ &= \int_{0}^{\pi} i \operatorname{Im}\{f(re^{i\theta})\} \biggl( \frac{z+re^{i\theta}}{z-re^{i\theta}} - \frac{z+re^{-i\theta}}{z-re^{-i\theta}} \biggr) \, \frac{\mathrm{d}\theta}{2\pi} \\ &= -\frac{2}{\pi} \int_{0}^{\pi} \frac{z}{r^2 + z^2 - 2rz \cos\theta} \, \varphi_r(\theta) \, \mathrm{d}\theta, \tag{*} \end{align*}
where $\varphi_r(\theta)$ is defined by
$$ \varphi_r(\theta) = r \operatorname{Im}\{f(re^{i\theta})\} \sin \theta $$
and the relation $\operatorname{Im}\{f(\overline{z})\} = -\operatorname{Im}\{f(z)\}$ is utilized in the second step. Now define
$$ C_r = \frac{2}{\pi} \int_{0}^{\pi} \varphi_r(\theta) \, \mathrm{d}\theta.$$
The assumption tells that $\varphi_r$ is non-negative, and so, $C_r \geq 0$. Moreover, $\text{(*)}$ applied to $z = R$ with a fixed $R > 0$ and $0 < r < R$ shows that
\begin{align*} C_r &= \frac{2(R + r)^2}{\pi R} \int_{0}^{\pi} \frac{R}{(R+r)^2} \, \varphi_r(\theta) \, \mathrm{d}\theta \\ &\leq \frac{2(R + r)^2}{\pi R} \int_{0}^{\pi} \frac{R}{r^2 + R^2 - 2rR \cos\theta} \, \varphi_r(\theta) \, \mathrm{d}\theta \\ &= \frac{(R + r)^2}{R} \left| h(R) - g(r^2/R) \right| \end{align*}
and so, $C_r$ is bounded as $r \to 0^+$. Finally,
\begin{align*} \left| z h(z) \right| &\leq \left| z g(r^2/z) \right| + \left| z ( h(z) - g(r^2/z)) \right| \\ &\leq \left| z g(r^2/z) \right| + \frac{2}{\pi} \int_{0}^{\pi} \frac{\left| z \right|^2}{\left|z\right|^2 - r^2 - 2r\left|z\right|} \, \varphi_r(\theta) \, \mathrm{d}\theta \\ &= \left| z g(r^2/z) \right| + \frac{\left| z \right|^2}{\left|z\right|^2 - r^2 - 2r\left|z\right|} C_r, \end{align*}
and so, taking limit superior as $r \to 0^+$ gives
$$ \left| z h(z) \right| \leq \limsup_{r\to 0^+} C_r < \infty. $$
Now this inequality holds for any $0 < |z| < 1$, and so, $zh(z)$ has a removable singularity at $0$ and therefore $f(z)$ cannot have an essential singularity at $0$.