# Derivative of Complex Polynomial

## Theorem

Let $a_n \in \C$ for $n \in \left\{ {0, 1, \ldots, N}\right\}$, where $N \in \N$.

Let $f: \C \to \C$ be a complex polynomial defined by $\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^N a_n z^n$.

Then $f$ is complex differentiable and its derivative is:

$\displaystyle f' \left({z}\right) = \sum_{n \mathop = 1}^N n a_n z^{n-1}$

## Proof

For all $n > N$, put $a_n = 0$.

Then:

$\displaystyle f \left({z}\right) = \sum_{n \mathop = 0}^\infty a_n z^n$

The result now follows from Derivative of Complex Power Series.

$\blacksquare$