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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

The square map sends the Cartesian grid lines off the coordinate axes to two orthogonal families of parabolas Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 55 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

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)