Derivative of Exponential Function/Complex

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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:

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

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:

\(\ds \map {D_z} {\exp z}\)

\(=\)

\(\ds \map {D_z} {\paren {1 + \dfrac z n}^n}\)

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.

$\blacksquare$