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 $\left \langle{E_n}\right \rangle$ defined by:

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

Then $\left \langle{E_n}\right \rangle$ is uniformly convergent on compact subsets of $\C$.

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

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

Hence:

\(\displaystyle D_z \left({\exp z}\right)\)

\(=\)

\(\displaystyle D_z \left({1 + \dfrac z n}\right)^n\)

Definition of Complex Exponential Function

\(\displaystyle \)

\(=\)

\(\displaystyle \lim_{n \to \infty} D_z \left({1 + \dfrac z n}\right)^n\)

Derivative of Sequence of Holomorphic Functions

\(\displaystyle \)

\(=\)

\(\displaystyle \lim_{n \to \infty} \left({ n \left({1 + \dfrac z n}\right)^{n - 1} \times \frac 1 n}\right)\)

Chain Rule

\(\displaystyle \)

\(=\)

\(\displaystyle \lim_{n \to \infty} \left({1 + \dfrac z n}\right)^{n - 1}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \lim_{n \to \infty} \left({ \left({1 + \dfrac z n}\right)^n \times \frac n {n + z} }\right)\)

\(\displaystyle \)

\(=\)

\(\displaystyle \left({ \lim_{n \to \infty} \left({1 + \dfrac z n}\right)^n }\right) \times \left({ \lim_{n \to \infty} \frac n {n + z} }\right)\)

Complex Derivative of Product is Product of Complex Derivative

\(\displaystyle \)

\(=\)

\(\displaystyle \exp z \times 1\)

Definition of Complex Exponential Function

Hence the result.

$\blacksquare$