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Spring 2013 - Problem 9

convex hull

Let $P\p{z}$ be a non-constant complex polynomial, all of whose zeroes lie in a half-plane $\set{z \in \C \mid \Re{z} < \sigma}$ . Show that all the zeroes of $P'\p{z}$ also lie in the same half-plane $\set{z \in \C \mid \Re{z} < \sigma}$ .

Hint: Compute the log-derivative $\frac{P'\p{z}}{P\p{z}}$ of $P$ .

Solution.

Write $P\p{z} = c \prod_{j=1}^n \p{z - a_j}$ , where $a_j$ are the roots of $P$ . Then

P'\p{z} = \sum_{j=1}^n \frac{P\p{z}}{z - a_j} \implies \frac{P'\p{z}}{P\p{z}} = \sum_{j=1}^n \frac{1}{z - a_j}.

Let $z_0$ be a root of $P'$ . If $z_0$ is a root of $P$ also, then clearly $z_0$ is in the same half-plane as the roots of $P$ . Otherwise, we have $0 = \sum_{j=1}^n \frac{1}{z_0 - a_j}$ . Taking real and imaginary parts, we see

0 = \sum_{j=1}^n \frac{\Re{z_0} - \Re{a_j}}{\abs{z_0 - a_j}^2} \quad\text{and}\quad 0 = \sum_{j=1}^n \frac{-\Im{z_0} + \Im{a_j}}{\abs{z_0 - a_j}^2}.

Let $\alpha = \sum_{j=1}^n \frac{1}{\abs{z_0 - a_j}^2}$ , and so rearranging yields

\begin{gathered} \Re{z_0} = \frac{1}{\alpha} \sum_{j=1}^n \frac{\Re{a_j}}{\abs{z_0 - a_j}^2} \quad\text{and}\quad \Im{z_0} = \frac{1}{\alpha} \sum_{j=1}^n \frac{\Im{a_j}}{\abs{z_0 - a_j}^2} \\ z_0 = \frac{1}{\alpha} \sum_{j=1}^n \frac{a_j}{\abs{z_0 - a_j}^2}. \end{gathered}

But observe that by definition, $\frac{1}{\alpha} \sum_{j=1}^n \frac{1}{\abs{z_0 - a_j}^2} = 1$ , i.e., $z_0$ is a convex combination of the roots of $P$ . Thus, $z_0$ lies in the convex hull of the roots of $P$ , and because $\set{\Re{z} < \sigma}$ is convex, it follows that $z_0$ lies in the same half-plane.