Prove that $0$ is not an essential singularity (UW Madison Qualifying exam) 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!!
 A: 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 $0$.
A: 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
A: The following stuff is based on @Sangchulee's ideas. Hopefully this works.
Let the Laurent expansion of $f$ at $0$ be $$f=\sum_{n=-\infty}^\infty a_nz^n,$$ define $$g=\sum_{n=0}^\infty a_{-n}z^n,$$ then $g$ is an entire function. Furthermore, there exists $a>1$ such that $|a_n|\leq a^n$ for all $n\in\mathbb{Z}^+$. Therefore, if $z\in\mathbb{C}$ and $|z|>a$,
then $$\bigg|\text{Im }g(z)-\text{Im }f\bigg(\frac{1}{z}\bigg)\bigg|\leq\frac{\frac{a}{|z|}}{1-\frac{a}{|z|}}\leq\frac{a}{|z|}.$$ Furthermore, $\overline{g(z)}=g(\overline{z})$ for each $z\in\mathbb{C}.$
If $z_0\in\mathbb{C}$, $R>0$ is a number such that $R>|z_0|$, the Schwartz integral formula implies that $$g(z_0)=\int_0^{2\pi}i\,\text{Im }g(Re^{i\theta})\frac{Re^{i\theta}+z_0}{Re^{i\theta}-z_0}\frac{d\theta}{2\pi}+K$$ for some $K\in\mathbb{R}$. Of course, we may assume that $K=0$ by subtracting $K$ from $g$.
Then $$g(z_0)=\int_0^{\pi}i\,\text{Im }g(Re^{i\theta})\frac{Re^{i\theta}+z_0}{Re^{i\theta}-z_0}\frac{d\theta}{2\pi}+\int_\pi^{2\pi}i\,\text{Im }g(Re^{i\theta})\frac{Re^{i\theta}+z_0}{Re^{i\theta}-z_0}\frac{d\theta}{2\pi}$$$$=\int_0^\pi i\,\text{Im }g(Re^{i\theta})\bigg(\frac{Re^{i\theta}+z_0}{Re^{i\theta}-z_0}-\frac{Re^{-i\theta}+z_0}{Re^{-i\theta}-z_0}\bigg)\,\frac{d\theta}{2\pi}$$$$=\int_0^\pi\frac{2}{\pi}\text{Im }g(Re^{i\theta})\frac{Rz_0\sin\theta}{R^2-2Rz_0\cos\theta+z_0^2}\,d\theta.$$
Take $z_0=a$, if $R>a$ is large enough, then $$|g(a)|=\bigg|\int_0^\pi\frac{2}{\pi}\text{Im }g(Re^{i\theta})\frac{Ra\sin\theta}{R^2-2Ra\cos\theta+a^2}\,d\theta\bigg|$$$$\geq \bigg|\int_0^\pi\frac{2}{\pi}\text{Im }f(R^{-1}e^{-i\theta})\frac{Ra\sin\theta}{R^2-2Ra\cos\theta+a^2}\,d\theta\bigg|-\int_0^{\pi}\frac{2}{\pi}\frac{a}{R}\frac{Ra\sin\theta}{R^2-2Ra-a^2}\,d\theta$$$$=\int_0^\pi\bigg|\frac{2}{\pi}\text{Im }f(R^{-1}e^{-i\theta})\frac{Ra\sin\theta}{R^2-2Ra\cos\theta+a^2}\bigg|\,d\theta-\frac{4a^2}{\pi(R^2-2Ra-a^2)}$$$$\geq \int_0^\pi\bigg|\frac{2}{\pi}\text{Im }f(R^{-1}e^{-i\theta})\frac{Ra\sin\theta}{R^2+2Ra+a^2}\bigg|\,d\theta-\frac{4a^2}{\pi(R^2-2Ra-a^2)}$$
Therefore, if $R>a$ is large enough, we have $$\int_0^{\pi}\bigg|\frac{2}{\pi}\text{Im }f(R^{-1}e^{-i\theta})R\sin\theta\bigg|\,d\theta\leq \bigg(g(a)+\frac{4a^2}{\pi(R^2-2Ra-a^2)}\bigg)\frac{(R+a)^2}{a}.$$
Hence $$\bigg|\frac{g(z_0)}{z_0}\bigg|=\bigg|\int_0^\pi\frac{2}{\pi}\text{Im }g(Re^{i\theta})R\sin\theta\frac{1}{R^2-2Rz_0\cos\theta+z_0^2}\,d\theta\bigg|$$$$\leq \frac{1}{R^2-2R|z_0|-|z_0|^2}\bigg(g(a)+\frac{4a^2}{\pi(R^2-2Ra-a^2)}\bigg)\frac{(R+a)^2}{a}$$$$+\int_0^\pi\frac{2}{\pi}\frac{a}{R}\frac{R\sin\theta}{R^2-2R|z_0|-|z_0|^2}\,d\theta$$$$=\frac{g(a)}{a}\frac{(R+a)^2}{R^2-2R|z_0|-|z_0|^2}+F(a,R,|z_0|),$$ where $F(a,R,|z_0|)$ is an expression that goes to $0$ if we take $R=3|z_0|$, then let $|z_0|$ goes to $\infty$. This is enough to show that $\big|\frac{g(z_0)}{z_0}\big|$ is bounded at infinity, so $g$ is a constant or an order 1 polynomial by Cauchy's integral formula, so $f$ has a simple pole or a removable singularity at $0$.
