How to show these complex conjugates of exponential functions

Question

I'm supposed to show that the $\overline{e^z}$ is equal to $e^{\displaystyle\overline z}$, and I'm not really sure where to start!

Any help would be really appreciated.

Answer 1

Here is a different approach from those in comments.

It's well known that $z\mapsto\overline{z}$ is a continuous, additive and multiplicative map. Thus \begin{equation} \begin{split} \overline{e^z} & = \overline{\sum_{n=0}^{\infty}\frac{z^n}{n!}} & = \overline{\lim_{n\to\infty} \sum_{k=0}^n \frac{z^k}{k!}} \\ & = \lim_{n\to\infty}\overline{\sum_{k=0}^n \frac{z^k}{k!}} & = \lim_{n\to\infty}\sum_{k=0}^n \frac{\overline{z}^k}{k!} \\ & = \sum_{n=0}^{\infty} \frac{\overline{z}^n}{n!} & = e^{\overline{z}}, \end{split} \end{equation} as desired.