Underrated Complex Analysis Theorem: Schwarz Lemma

The Schwarz Lemma is a relatively basic lemma in Complex Analysis, that can be said to be of greater importance that it seems. There is a whole article written on it.

The conditions and results of Schwarz Lemma are rather difficult to memorize offhand, some tips I gathered from the net on how to memorize the Schwarz Lemma are:

Conditions: f:D\to D holomorphic and fixes zero.

Result 1: |f(z)|\leq|z| can be remembered as “Range of f” subset of “Domain”.

|f'(0)|\leq 1 can be remembered as some sort of “Contraction Mapping”.

Result 2: If |f(z)|=|z|, or |f'(0)|=1, then f=az where |a|=1. Remember it as “f is a rotation”.

If you have other tips on how to remember or intuitively understand Schwarz Lemma, please let me know by posting in the comments below.

Finally, we proceed to prove the Schwarz Lemma.

Schwarz Lemma

Let D=\{z:|z|<1\} be the open unit disk in the complex plane \mathbb{C} centered at the origin and let f:D\to D be a holomorphic map such that f(0)=0.

Then, |f(z)|\leq |z| for all z\in D and |f'(0)|\leq 1.

Moreover, if |f(z)|=|z| for some non-zero z or |f'(0)|=1, then f(z)=az for some a\in\mathbb{C} with |a|=1 (i.e.\ f is a rotation).

Proof

Consider g(z)=\begin{cases}  \dfrac{f(z)}{z} &\text{if }z\neq 0,\\  f'(0) &\text{if }z=0.  \end{cases}
Since f is analytic, f(z)=0+a_1z+a_2z^2+\dots on D, and f'(0)=a_1. Note that g(z)=a_1+a_2z+\dots on D, so g is analytic on D.

Let D_r=\{z:|z|\leq r\} denote the closed disk of radius r centered at the origin. The Maximum Modulus Principle implies that, for r<1, given any z\in D_r, there exists z_r on the boundary of D_r such that \displaystyle |g(z)|\leq|g(z_r)|=\frac{|f(z_r)|}{|z_r|}\leq\frac{1}{r}.

As r\to 1 we get |g(z)|\leq 1, thus |f(z)|\leq|z|. Thus
\begin{aligned}  |f'(0)|&=|\lim_{z\to 0}\frac{f(z)}{z}|\\  &=\lim_{z\to 0}|\frac{f(z)}{z}|\\  &\leq1.  \end{aligned}
Moreover, if |f(z)|=|z| for some non-zero z\in D or |f'(0)|=1, then |g(z)|=1 at some point of D. By the Maximum Modulus Principle, g(z)\equiv a where |a|=1. Therefore, f(z)=az.

Rouche’s Theorem and Applications

This blog post is on Rouche’s Theorem and some applications, namely counting the number of zeroes in an annulus, and the fundamental theorem of algebra.

Rouche’s Theorem: Let f(z), g(z) be holomorphic inside and on a simple closed contour K, such that |g(z)|<|f(z)| on K. Then f and f+g have the same number of zeroes (counting multiplicities) inside K.

Rouche’s Theorem is useful for scenarios like this: Determine the number of zeroes, counting multiplicities, of the polynomial f(z)=2z^5-6z^2-z+1=0 in the annulus 1\leq |z|\leq 2.

Solution:

Let K_1 be the unit circle |z|=1. We have

\begin{aligned}|2z^5-z+1|&\leq |2z^5|+|z|+|1|\\    &=2+1+1\\    &=4\\    &<6\\    &=|-6z^2|    \end{aligned}

on K_1.

Since -6z^2 has 2 zeroes in K_1, therefore f has 2 zeroes inside K_1, by Rouche’s Theorem.

Let K_2 be the circle |z|=2

\begin{aligned}    |-6z^2-z+1|&\leq |-6z^2|+|-z|+|1|\\    &=6(2^2)+2+1\\    &=27\\    &<64\\    &=|2z^5|    \end{aligned}

on K_2. Therefore f has 5 zeroes inside K_2.

Therefore f has 5-2=3 zeroes inside the annulus.

We do a computer check using Wolfram Alpha (http://www.wolframalpha.com/input/?i=2z%5E5-6z%5E2-z%2B1%3D0). The moduli of the five roots are (to 3 significant figures): 0.489, 0.335, 1.46, 1.45, 1.45. This confirms that 3 of the zeroes are in the given annulus.

Fundamental Theorem of Algebra Using Rouche’s Theorem

Rouche’s Theorem provides a rather short proof of the Fundamental Theorem of Algebra: Every degree n polynomial with complex coefficients has exactly n roots, counting multiplicities.

Proof: Let f(z)=a_0+a_1z+a_2z^2+\dots+a_nz^n. Chose R\gg 1 sufficiently large so that on the circle |z|=R,

\begin{aligned} |a_0+a_1z+a_2z^2+\dots+a_{n-1}z^{n-1}|&\leq|a_0|+|a_1|R+|a_2|R^2+\dots+|a_{n-1}|R^{n-1}\\    &<(\sum_{i=0}^{n-1}|a_i|)R^{n-1}\\    &<|a_n|R^n\\    &=|a_nz^n|    \end{aligned}

Since a_nz^n has n roots inside the circle, f also has n roots in the circle, by Rouche’s Theorem. Since R can be arbitrarily large, this proves the Fundamental Theorem of Algebra.