# Derivative of Exponential Function/Complex

## Theorem

The complex exponential function is its own derivative.

That is:

$D_z \left({\exp z}\right) = \exp z$

## Proof from Sequence Definition

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

$\displaystyle E_n \left( {z} \right) = \left({1 + \dfrac z n}\right)^n$

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

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

Hence:

 $\displaystyle D_z \left({\exp z}\right)$ $=$ $\displaystyle D_z \left({1 + \dfrac z n}\right)^n$ $\quad$ Definition of Complex Exponential Function $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} D_z \left({1 + \dfrac z n}\right)^n$ $\quad$ Derivative of Sequence of Holomorphic Functions $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} \left({ n \left({1 + \dfrac z n}\right)^{n - 1} \times \frac 1 n}\right)$ $\quad$ Chain Rule $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} \left({1 + \dfrac z n}\right)^{n - 1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \lim_{n \to \infty} \left({ \left({1 + \dfrac z n}\right)^n \times \frac n {n + z} }\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({ \lim_{n \to \infty} \left({1 + \dfrac z n}\right)^n }\right) \times \left({ \lim_{n \to \infty} \frac n {n + z} }\right)$ $\quad$ Complex Derivative of Product is Product of Complex Derivative $\quad$ $\displaystyle$ $=$ $\displaystyle \exp z \times 1$ $\quad$ Definition of Complex Exponential Function $\quad$

Hence the result.

$\blacksquare$