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Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero

Statement

Let $P (z) = \sum_{k = 0}^{n} a_{k} z^{k}$ be a complex polynomial. Then $P$ is entire and

$P^{'} (z) = \sum_{k = 1}^{n} k a_{k} z^{k - 1} .$

This includes the zero polynomial and constant polynomials, whose derivative is zero. If $P, Q$ are complex polynomials, then the set $D = {z \in C : Q (z) \neq 0}$ is open, $P / Q$ is holomorphic on $D$ , and

$(P / Q)^{'} = \frac{P^{'} Q - P Q^{'}}{Q^{2}} .$

When $Q$ is a nonzero constant, $D = C$ ; when $Q$ is the zero polynomial, $D = \emptyset$ and no rational function is defined there.

Facts & Assumptions

Given: Complex polynomials $P, Q$ with finite coefficient support.

[L1]

Constants and the identity have derivatives $0$ and $1$ ; finite linear combinations, products, reciprocals, and quotients obey the displayed derivative rules wherever denominators are nonzero (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[F1]

A polynomial over a commutative ring is a finitely supported coefficient sequence, written formally as a finite sum $\sum_{i} a_{i} x^{i}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L2]

A complex-differentiable function is continuous at each point of differentiability (Complex differentiability at a point implies continuity there).

[L3]

Complex modulus is definite and subadditive (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

technique · direct

1.1

For $m \geq 1$ and $h \neq 0$ , the factorization $((z + h)^{m} - z^{m}) / h = \sum_{j = 0}^{m - 1} (z + h)^{m - 1 - j} z^{j}$ has limit $m z^{m - 1}$ ; for $m = 0$ the function is constant and has derivative $0$ by [L1].

L1algebra

2.1

By finite support [F1], $P$ is a finite linear combination of these powers. The linearity rule [L1] and step 1.1 make $P$ entire with the asserted derivative, including the empty-support zero polynomial.

step 1.1F1L1

3.1

Fix $z_{0} \in D$ . By step 2.1 and [L2], $Q$ is continuous at $z_{0}$ , so some neighbourhood satisfies $∣ Q (z) - Q (z_{0}) ∣ < ∣ Q (z_{0}) ∣$ ; [L3] then forces $Q (z) \neq 0$ . Thus $D$ is open.

step 2.1L2L3given

4.1

On $D$ , both polynomials are holomorphic and $Q$ is nonzero. The quotient rule [L1] gives the displayed derivative. If $Q$ is a nonzero constant then it never vanishes, while for $Q = 0$ the set $D$ is empty.

step 2.1step 3.1L1algebra∎

Depends on

Used by

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Sources

J. Lebl, Guide to Cultivating Complex Analysis, Proposition 2.2.5 (standard reference, not scraped)
R. Howell and J. Mathews, Complex Analysis, §3.1 (standard reference, not scraped)