Derivative of Exponential Function/Complex

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Theorem

That is:

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

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 $\C$.

Hence:

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

$\ds \exp z \times 1$

Hence the result.

$\blacksquare$