Show that $\exp(\overline{z}) = \overline{\exp(z)} $

Question

Given that $z \in \mathbb{C}$, I want to show that $\exp(\overline{z}) = \overline{\exp(z)} $. For LHS I can see that we will get some real number as $\overline{z}$ is some real number. But should RHS be a complex number which is a complex conjugate of $e^z$? If so how can this equality hold?

Answer 1

Using the expansion of $z$ as a slightly different way of obtaining the answer, write $$ e^z = \sum_{k=0}^\infty \frac{1}{k!}\;z^k $$ Thus \begin{align} \overline{e^z} &= \overline{\sum_{k=0}^\infty \frac{1}{k!}\;z^k} \\ &= \sum_{k=0}^\infty \overline{\;\frac{1}{k!}\;}\;\overline{z^k}\\ &= \sum_{k=0}^\infty \frac{1}{k!}\;{\overline{z}\;}^k\\ &= e^{\overline{z}} \end{align}

Answer 2

Another proof.
$$ e^z\quad\text{and}\quad\overline{\;e^{\overline{z}}\;} $$ are two analytic functions that agree on the real line. By the identity theorem, they agree everywhere.