Derivative of Exponential Function/Complex

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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:

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

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:

\(\displaystyle D_z \left({\exp z}\right)\) \(=\) \(\displaystyle D_z \left({1 + \dfrac z n}\right)^n\) $\quad$ 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.

$\blacksquare$