# Read Lucas's Theorem

Suppose that $P(z)$ is a polynomial in the complex plane. The theorem says that all zeros of $P’(z)$ are inside a half plane in which all zeros of $P(z)$ lie.

Ahlfors says that a directed line $z = a + bt$ determines a right half plane consisting of all points with $\Im(z - a) / b < 0$… (see Chap. 1, Sec. 2.3)1 After drawing the figure for the drawing the figure for $z = -(1 + i)t$, I realized that I should pay attention to the word “directed” and interpret “right” as “to the right of the vector $b$ drawn on the line $z = a + bt$”.

The following equation puzzled me for a while. (see Chap. 2, Sec. 1.3)2

$$\Im\left( \frac{z - \alpha_k}{b} \right) = \Im\left( \frac{z - a}{b} \right) - \Im\left( \frac{\alpha_k - a}{b} \right) > 0 \label{stuck}$$

I tried sketching a diagram to understand what’s going on, but this isn’t so helpful. In fact, the above equation starts from $z - \alpha_k = (z - a) - (\alpha_k - a)$.

The whole proof makes use of a half plane $H := \{z \in \C \mid \Im[(z - a) / b] < 0\}$ in which all zeros $\alpha_1,\ldots,\alpha_n$ of $P(z)$ lie, and it follows the logic of proof by contradiction: each $\alpha_k$ is assumed to be in $H$ while $z$ isn't. It ends with the conclusion that $\Im(b P'(z) / P(z)) < 0$ by \eqref{stuck} and the equation

$\frac{P'(z)}{P(z)} = \sum_{k = 1}^n \Im\left( \frac{1}{z - \alpha_k} \right).$

Finally, I saw a corollary of this theorem: all zeros of $P’(z)$ are inside the smallest convex polygon in which all zeros of $P(z)$ lie.

1. Ahlfors, L. (1979). Complex analysis. Auckland: McGraw-Hill.

2. ibid