📂Complex Anaylsis

Ernestrom-Kakeya Theorem Proof

Theorem ¹

Let $\left\{ a_{i} \right\}_{i=0}^{n} \subset \mathbb{R}$ such that $a_0 > a_1 > \cdots > a_n > 0$. Then for the polynomial function $$P(z) := a_0 + a_1 z + \cdots + a_{n-1} z^{n-1} + a_n z^n$$ all roots $z \in \mathbb{C}$ satisfy $|z| \ge 1$.

Proof

If there is a root of $P(z) = 0$ at $z=1$, then we have $\displaystyle 0 = P(1) = \sum_{i=0}^{n} a_{i} > 0$, so the root must be $z \ne 1$. Multiply both sides of the equation $P(z) = 0$ by $z$ and subtract from the original equation to express $a_0$ as $$a_0 = (1-z)P(z) + (a_0 - a_1) z + \cdots + (a_{n-1} - a_n) z^n + a_n z^{n+1}$$ If we assume that a root $z \ne 1$ of $P(z) = 0$ satisfies $|z| < 1$ given that $a_0 > a_1 > \cdots > a_n > 0$, then we have $$\begin{align*} & |a_0| < |(1-z)P(z)| + (a_0 - a_1) + \cdots + (a_{n-1} - a_n) + a_n \\ \implies& |a_0| < |(1-z)P(z)| + a_0 + (- a_1 + a_1) + \cdots + (- a_{n-1} + a_{n-1} )+ (- a_n + a_n ) \\ \implies& a_0 = |a_0| < |(1-z)P(z)| + a_0 \\ \implies& 0 < |(1-z)P(z)| \end{align*}$$ Yet, since we assumed $z \ne 1$ is a root of $P(z) = 0$, we find a contradiction $$0 < |(1-z)P(z)| = 0$$ This indicates the wrongness of the assumption that $| z | < 1$, hence we must have $|z | \ge 1$.

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