# Derivative of Exponential Function/Complex

## Theorem

The complex exponential function is its own derivative.

That is:

$\map {D_z} {\exp z} = \exp z$

## Proof from Sequence Definition

Take the definition of $\exp$ to be the limit of the sequence $\sequence {E_n}$ defined by:

$\ds \map {E_n} z = \paren {1 + \dfrac z n}^n$

Further, $\C$ is an open, connected subset of $\C$.

So the hypotheses of Derivative of Sequence of Holomorphic Functions are satisfied.

Hence:

 $\ds \map {D_z} {\exp z}$ $=$ $\ds \map {D_z} {\paren {1 + \dfrac z n}^n}$ Definition of Complex Exponential Function $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \map {D_z} {\paren {1 + \dfrac z n}^n}$ Derivative of Sequence of Holomorphic Functions $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \paren {n \paren {1 + \dfrac z n}^{n - 1} \times \frac 1 n}$ Chain Rule $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^{n - 1}$ $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \paren {\paren {1 + \dfrac z n}^n \times \frac n {n + z} }$ $\ds$ $=$ $\ds \paren {\lim_{n \mathop \to \infty} \paren {1 + \dfrac z n}^n} \times \paren {\lim_{n \mathop \to \infty} \frac n {n + z} }$ Complex Derivative of Product is Product of Complex Derivative $\ds$ $=$ $\ds \exp z \times 1$ Definition of Complex Exponential Function

Hence the result.

$\blacksquare$