TITLE: Prove that $0$ is not an essential singularity (UW Madison Qualifying exam)
QUESTION [5 upvotes]: 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!!

REPLY [2 votes]: This problem (the part with simple pole or analytic) has an elementary solution since first $f'(x) \ne 0$ if $x \in (-1,0) \cup (0,1)$ by the local form of an analytic function (if $f'(r)=0, a_n,  n \ge 2$ is the first non-zero coefficient at $r, f(z)=f(r)+a_n(z-r)^n+O((z-r)^{n+1}), a_n \in \mathbb R$ and $(z-r)^n$ maps each half circle centered at $r$ on the full circle, which means that $f$ does so near $r$ and that contradicts the hypothesis on $\Im f$)
But then $f$ is monotonic on $(-1,0)$ and on $(0,1)$ which means that $f$ has limits at zero from both left and right (possibly $\pm \infty$ of course) so for $r>0$ small, $\mathbb R -(f(-r,0) \cup f(0,r))$ contains a non-degenerate interval $I$. But this means $\mathbb C - f(D(0,r)^*)$ contains $I$ too and that shows that $0$ cannot be an essential singularity (directly without Picard by mapping $\mathbb C-I$ into the unit disc with a meromorphic $g$ etc)
But then if $0$ is removable, $f'(0) \ne 0$ by the same argument as for $f'(r)$ above, while if $0$ is a pole, the local form of it shows again it must be simple (same reason that $1/z^n, n \ge 2$ scrambles imaginary parts) with negative residue.
Note that $f$ may have zeroes on the real axis (at most two of course by monotonicity) so one cannot apply directly the removable singularity result to $-1/f$ though indeed one can restrict to a neighborhood of zero where $f$ is non-zero and do it there since of course, the same arguments (showing that it has a simple pole or a removable singularity) apply if $f$ is typically real on a small punctured disc only