Suppose $f$ is non-constant and holomorphic in a neighborhood of the closed unit disk, s.t. $|f(z)| = 1$ for all $|z| = 1$. Then show that as $f(e^{i\theta})$ traverses the unit circle and makes at least one counterclockwise loop.

There was also the following hint: For each point on the unit circle, we can find a neighborhood on which $\log f$ is analytic. Use maximum modulus principle.

I attempted to follow the hint: By maximum modulus principle, we know that $|f(z)| < 1$ anywhere inside the open unit disk. Given any point on the unit circle, we can choose a neighborhood of that point so that modulus of $f$ in that neighborhood is nearly $1$ throughout. From there, we can apply the maximum modulus principle on this neighborhood to show that $\arg f(z)$ achieves a maximum on the boundary of this neighborhood. I don't know what to do from there.

Any help on this question would be greatly appreciated.

EDIT: I found this thread from a while ago that has a comment claiming that such a function must be a Blaschke product, but I can't seem to follow his logic; he uses a multiplicity argument, but I don't know how to use this if $f$ has an infinite number of zeros in the unit disk.

EDIT 2: I now know that $f$ has a finite number of zeros, but I still do not know how to proceed.

EDIT 3: This is a solution that utilizes the Argument Principle and Winding numbers. However, since we have not yet covered these topics $\textbf{I am still looking for another solution}$. NOTE: the solution posted by Daniel Fischer below is exactly what I was looking for, and solves the problem without argument principle and winding numbers.

Let $f$ be a function satisfying the hypotheses of the problem. Suppose $\frac{1}{f(z)}$ is holomorphic in the unit disk. Then by maximum modulus principle, we know that $\frac{1}{f(z)} < 1$ for all $z$ in the interior of the disk. This is a contradiction, since we know that $f(z) < 1$ for all $z$ in the interior of the disk by maximum modulus principle on the original function. Thus $\frac{1}{f(z)}$ is not analytic, i.e. $f$ has at least one zero in the interior of the disk.

Let $\gamma(\theta) = e^{i\theta}$, a parameterization of the unit circle, traveling counter clockwise. Then the winding number of $f\circ \gamma$ is \begin{align*} \frac{1}{2\pi i }\int\limits_{f\circ \gamma} \frac{1}{z}dz &= \frac{1}{2\pi i }\int\limits_0^{2\pi} \frac{f'(e^{i\theta}) i e^{i\theta} }{f(e^{i\theta})}d\theta \\ &= \frac{1}{2\pi i }\int\limits_{\gamma} \frac{f'(z)}{f(z)} dz \\ &= \text{Number of zeros of }f - \text{ number of poles of } f \\ &= \text{Number of zeros of }f \\ &\geq 1 \end{align*}

Thus, $f(e^{i\theta})$ goes counterclockwise around the unit disk at least once.

  • $\begingroup$ If $\lvert f(z)\rvert = 1$ for all $z$ on the unit circle, then $f$ can have only finitely many zeros in the unit disk. All that you need is that $f$ has at least one zero in the unit disk. $\endgroup$ Sep 21, 2014 at 20:32
  • 1
    $\begingroup$ And, the big question is: What has $\frac{f'(z)}{f(z)}$ to do with it? $\endgroup$ Sep 21, 2014 at 20:33
  • 1
    $\begingroup$ It seems you haven't yet learned the residue theorem, winding numbers, or the argument principle? $\endgroup$ Sep 21, 2014 at 20:49
  • 1
    $\begingroup$ Indeed it is pretty easy to show that $f$ must have at least one zero in the unit disk. If you don't find the solution yourself, be invited to ask for a hint. So, without using the theory, getting the result elementarily: the maximum principle says (here) that $\lvert z\rvert < 1 \implies \lvert f(z)\rvert < 1$. What does that imply about the direction of the change of $\arg f(z)$ in each point on the unit circle? $\endgroup$ Sep 21, 2014 at 21:09
  • 1
    $\begingroup$ Oy, you wouldn't have needed to place a bounty. I had already started writing the answer, but my bath was ready, so I couldn't finish it. $\endgroup$ Sep 21, 2014 at 22:39

1 Answer 1


The local change of $\arg f(z)$ along the unit circle is given by

$$\begin{align} \frac{\partial}{\partial\varphi} \arg f\left(e^{i\varphi}\right) &= \frac{\partial}{\partial\varphi} \frac{1}{i}\left(\log \left\lvert f\left(e^{i\varphi}\right)\right\rvert + i \arg f\left(e^{i\varphi}\right)\right)\\ &= \frac{1}{i}\frac{\partial}{\partial\varphi} \log f\left(e^{i\varphi}\right)\\ &= \frac{1}{i} \frac{f'\left(e^{i\varphi}\right)}{f\left(e^{i\varphi}\right)} \frac{\partial e^{i\varphi}}{\partial \varphi}\\ &= \frac{e^{i\varphi}f'\left(e^{i\varphi}\right)}{f\left(e^{i\varphi}\right)} \end{align}$$

since $\left\lvert f\left(e^{i\varphi}\right)\right\rvert \equiv 1$ and thus $\log \left\lvert f\left(e^{i\varphi}\right)\right\rvert \equiv 0$.

So if we can show that $\frac{z f'(z)}{f(z)}$ is a non-negative real number on the unit circle, and non-zero except at possibly finitely many points, we have shown that $\arg f\left(e^{i\varphi}\right)$ is strictly increasing, and since the image of the unit circle under $f$ is a closed curve, the total change of the argument must then be a positive multiple of $2\pi$.

By the maximum principle, we have $\lvert z\rvert < 1\implies \lvert f(z)\rvert < 1$. Now consider a point $z_0$ on the unit circle with $f'(z_0) \neq 0$. Consider a path on the radius through $z_0$, $\gamma_r \colon t \mapsto (1-t)\cdot z_0$, and a path on the unit circle through $z_0$, $\gamma_c\colon \varphi \mapsto e^{i\varphi}\cdot z_0$. Composing these paths with $f$, we obtain one path $f\circ \gamma_r$ that enters the unit disk through $f(z_0)$, and one path $f\circ \gamma_c$ that lies on the unit circle and passes through $f(z_0)$. Thus $(f\circ \gamma_c)'(0) = f'(z_0)\cdot iz_0$ is tangent to the unit circle [more precisely, the line $\{f(z_0) + \lambda\cdot (f\circ \gamma_c)'(0) : \lambda\in \mathbb{R}\}$ is the tangent to the unit circle through $z_0$]. But the tangent is also the line $\{f(z_0) + \lambda\cdot i f(z_0) : \lambda\in \mathbb{R}\}$, hence $\frac{z_0f'(z_0)}{f(z_0)}$ is real.

And $(f\circ \gamma_r)'(0) = f'(z_0)\cdot (-z_0) = i\cdot (f\circ \gamma_c)'(0)$ points inside the unit circle. That tells you which sign $\frac{z_0f'(z_0)}{f(z_0)}$ has.


You must log in to answer this question.