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.


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 ( 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.


About mathtuition88
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