# Problem of the Week #17 - September 24th, 2012

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#### Chris L T521

##### Well-known member
Staff member
Here's this week's problem.

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Problem: Let $p(z)$ be a polynomial of degree $n$ that is nonzero on the unit circle $\mathbb{T}=\{z:|z|=1\}$. Define $m_0$ by the formula
$m_0=\frac{1}{2\pi i}\int_{\mathbb{T}}\frac{p^{\prime}(z)}{p(z)}\,dz.$
Show that $m_0\in\{0,\ldots,n\}$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
No one answered this question correctly (good attempt dwsmith, but your conclusion wasn't correct). You can find my solution below.

Proof: This problem becomes rather straight forward when you apply the argument principle. Since $p(z)$ is a polynomial, it is holomorphic (and meromorphic) in the interior of $\mathbb{T}$. By the fundamental theorem of algebra, $p(z)$ has exactly $n$ zeros, and has at most $n$ zeros in the interior of $\mathbb{T}$. Therefore, by the argument principle, we have
$m_0 = \int_{\mathbb{T}}\frac{p^{\prime}(z)}{p(z)}\,dz = \{\text{# of zeros p(z) has inside \mathbb{T}}\}-\{\text{# of poles p(z) has inside \mathbb{T}}\} \leq n-0 = n.$

It now follows that $0\leq m_0\leq n\implies m_0\in\{0,\ldots,n\}$. Q.E.D.

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