2.3Polynomials

Updated 25 Jun 2026Chapter (PDF)

Note that all polynomials are holomorphic. For now, we will accept the Fundamental Theorem of Algebra without proof and note that any polynomial $p$ over $\bC$ takes the unique form

p(z) = c(x - a_1)\cdots(x - a_n),

for not necessarily distinct $a_1, \dots, a_n$ . We say each $a_k$ is a zero of $p$ of order $h$ , where $h$ is the number of times $a_k$ coincides with the other roots.

Note that the order of a root can be determined by calculus. If $\zeta$ is a root of $p(z)$ of order $n$ , then note

p(\zeta) = p'(\zeta) = \cdots = p^{(n - 1)}(\zeta) = 0,

whereas $p^{(n)}(\zeta) \ne 0$ . In other words, the order of $\zeta$ is $n$ if and only if $p^{(n)}(\zeta)$ is the first nonvanishing derivative at $\zeta$ .

[2.3.1]Theorem(Gauss-Lucas Theorem)#

Suppose a convex hull is defined by $p(z)$ 's roots in the complex plane. Then $p'(z)$ has all its roots contained in the convex hull.

Proof.

Suppose $z_0$ is a root of $p'$ . If $z_0$ is a root of $p$ , then we are done—otherwise, note $p(z_0) \ne 0$ . Then

0 = \frac{p'(z_0)}{p(z_0)} = \sum_{k = 1}^n \frac{1}{z_0 - \lambda_k} = \sum_{k = 1}^n \frac{z_0 - \lambda_k}{|z_0 - \lambda_k|^2} = \sum_{k = 1}^n m_k(z_0 - \lambda_k),

where $\lambda_1, \dots, \lambda_n$ are the roots of $p$ and $m_k = 1/|z_0 - \lambda_k|^2 \in \bR$ . Rewriting yields

z_0 = \frac{\sum_k m_k \lambda_k}{\sum_k \lambda_k},

a weighted linear combination of the roots of $p$ (the points that define the hull), completing the proof.