Ngô Quốc Anh

July 18, 2009

Schwarz Reflection Principle and several applications, 2

I want to continue the topic “Schwarz Reflection Principle and several applications”. Today we discuss the following.

Question. If entire function g satisfies |g(z)|=1 whenever |z|=1, then there exist a non-negative integer number n and a constant c satisfying |c| = 1 such that g(z) = cz^n.

Solution. Suppose g has a zero at the origin of order n for some n\ge 0 and the other zeros of g in the unit disk are listed as a_1,a_2,\dots,a_m, repeating as necessary to account for multiplicities. The set of zeros in the unit disk must be finite since that set cannot have any limit points.


\displaystyle B(z) = z^n\prod_{j = 1}^m\frac {z - a_j}{1 - \overline{a_j}z}.

Note that |B(z)| = 1 for all |z| = 1. Let f(z) = \frac {g(z)}{B(z)}. Then f is an entire function with no zeros in the unit disk and with the property that |f(z)| = 1 for |z| = 1.

By the maximum modulus principle |f(z)| and \frac {1}{|f(z)|} must both achieve their maximum values for the unit disk on the boundary, from which we conclude that |f(z)| = 1 for all |z|\le 1. But then, that means that f(z) = c must be constant, with |c| = 1.

So we have that

\displaystyle g(z) = cB(z) = cz^n\prod_{j = 1}^m\frac {z - a_j}{1 - \overline{a_j}z}.

But that expression has poles whenever z = \frac1{\overline{a_j}}. The only way we could have g be an entire function is for there to be no such a_j. Hence we conclude that g(z) = cz^n for some nonnegative n.

Here is an even further simplification.

Dividing by some finite power of z (which does not change the property), we can assume that g(0)\ne 0. Now consider

\displaystyle \frac1{g(1/z)} \quad \text{ and } \quad \overline{g(\bar z)}.

They are meromorphic on \mathbb C and coincide on the unit circle. Thus they coincide everywhere. Hence

\displaystyle g(z)\to \frac1{g(0)} \quad \text{ as } \quad z\to\infty,

so g is bounded and, thereby, constant.

I will provide another proof mainly based on the Schwarz Reflection Principle. However, to this purpose, an extension of Morera’s Theorem for toy contours should be introduced firstly.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

Create a free website or blog at

%d bloggers like this: