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!!

  • $\begingroup$ re edit, $f(x)=e^{-1/x}$ has $f(-1/i\pi)=-1$ $\endgroup$ Jul 6, 2020 at 1:02
  • $\begingroup$ Not related to the essential singularity, but you may resolve the case of pole by an easier way: if $f$ has a pole at $0$, then consider $-1/f$ instead and apply the conclusion for the case of removable singularity. And I am also deeply interested in your question, too! $\endgroup$ Jul 6, 2020 at 1:53
  • $\begingroup$ @CalvinKhor You're right. My counterexample is false. $\endgroup$ Jul 6, 2020 at 2:41
  • $\begingroup$ Actually, would you mind enlightening me as to why your $g$ satisfies $\operatorname{Im}g(z) > 0$ (or even $\geq$) when $\operatorname{Im}(z)>0$? $\endgroup$ Jul 6, 2020 at 19:44
  • $\begingroup$ @SangchulLee At first I thought for $|z|$ large this is true. Then I realized my mistake. $\endgroup$ Jul 6, 2020 at 19:51

3 Answers 3


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 $0$.

  • $\begingroup$ Thanks! I will try to understand your proof. This is a 1997 qualifying exam problem, so the professors at Madison probably have a simple proof. $\endgroup$ Jul 6, 2020 at 5:49
  • $\begingroup$ @Simplyorange, On a second though, I think I can remove the measure-theoretic flavor and modify the proof accordingly. Let me try this. $\endgroup$ Jul 6, 2020 at 6:08
  • $\begingroup$ Thanks! I saw that you were a grad student at UCLA! I'm an incoming grad student at UCLA! $\endgroup$ Jul 6, 2020 at 6:16
  • $\begingroup$ @Simplyorange, That's great, congratulation! I hope you enjoy your incoming life at LA and UCLA, and most of all, stay safe. $\endgroup$ Jul 6, 2020 at 6:35
  • $\begingroup$ I think I understand your proof now! Very nice. Although I think you meant to use $h(1/z)$ instead of things like $g(r^2/z)$? I'll write a proof based on your ideas in the problem itself. $\endgroup$ Jul 6, 2020 at 19:11

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

  • $\begingroup$ (+1) Brilliant! $\endgroup$ Jul 7, 2020 at 3:09
  • $\begingroup$ @Sangchul - appreciate the nice comment but frankly I think your solution is really cool and interesting; this is just standard geometric function theory (typically real functions have some similar properties to univalent functions with real coefficients, so it stands to reason that in a punctured disc they either extend or have a simple pole) $\endgroup$
    – Conrad
    Jul 7, 2020 at 3:17
  • $\begingroup$ Still it is cool to learn such an argument :) Even I cannot tell my solution is purely original since it is motivated by a general theory of Nevanlinna-Herglotz functions (holomorphic functions from the upper half plane $\mathbb{H}$ to itself), although my knowledge on it is rather tangential. $\endgroup$ Jul 7, 2020 at 3:48

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 $0$.

  • $\begingroup$ Indeed the proof seems correct. Here are a couple of comments: (1) There are some sign mistakes. For instance, $$R^2-2Ra\color{red}{-a^2}.$$ (2) In deriving the key bound (the second to the last formula), It might be easier to start from $$-\operatorname{Im}f(R^{-1}e^{-i\theta})=|\operatorname{Im}f(R^{-1}e^{-i\theta})|\leq|\operatorname{Im}g(Re^{i\theta})|+\frac{a}{R}.$$ This would emphasize the role of the assumption on $f$. (3) In the last formula, I guess what you intended is $R^2-2R|z_0|+|z_0|^2$ instead of $R^2-2Rz_0-z_0^2$. (4) "... let $|z_0|$ goes to zero." Do you mean infinity? $\endgroup$ Jul 7, 2020 at 1:49
  • $\begingroup$ Yes! Thank you! I meant "letting $|z_0|$ goes to $\infty$". As for the signs, yes, it looks nicer if we write $R^2-2Ra+a^2$, I wrote $R^2-2Ra-a^2$ just to be extra cautious. I also wrote $R^2-2R|z_0|-|z_0|^2$ just to be extra cautious, since we only care about large $R$. Because $z_0$ is a complex number, I didn't check if it is still true that $|R^2-2Rz_0\cos\theta+z_0^2|\geq|R^2|-2R|z_0|+|z_0|^2$. $\endgroup$ Jul 7, 2020 at 2:21
  • $\begingroup$ That totally makes sense. Now it seems my turn to worry about whether the inequality $$|R^2-2Rz_0\cos\theta+z_0^2|\stackrel{?}\geq R^2-2R|z_0|+|z_0|^2$$ is true. In retrospect, it feels quite strange that I believed this inequality without any doubt... $\endgroup$ Jul 7, 2020 at 3:01
  • $\begingroup$ @SangchulLee Another person conrad just pointed out $0$ is not an essential singularity because $f'(z)>0$ if $z\in(-1,0)\cup(0,1)$, so $f$ is an increasing function on $(-1,0)$ and $(0,1)$, but $f$ is not real if $z$ is not real, this contradicts the Great Picard theorem, since $f$ cannot attain any real number infinitely many times near $0$. (I feel so dumb now, and I think I'll accept his/her answer instead since this is the solution UW Madison professors probably wanted.) $\endgroup$ Jul 7, 2020 at 3:04
  • $\begingroup$ I just saw that too, and even though I suspected the use of Great Picard Theorem at some point, I could never made it. How he makes use of that theorem is really brilliant. I also feel dumb for missing such an elegant argument. $\endgroup$ Jul 7, 2020 at 3:07

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