Why is $ \overline{e^z} = e^\overline{z} $?

Question

How can you conjugate an entire function? $ \overline{exp(z)} $ I need an equivalent.

I thought this is only possible with complex numbers.

What is the proof for $ \overline{e^z} = e^\overline{z} $ ? (Please don't involve a power series here.)

Answer 1

$$\overline{e^z}=\overline{e^xe^{iy}}=e^x(\overline{\cos y+i\sin y})=e^x(\cos y-i\sin y)=e^x(\cos(-y)+i\sin(-y))=e^{x-iy}=e^{\overline{z}}$$

Answer 2

Using another definition. $$ e^z = \sum_{k=0}^\infty \frac{1}{k!}\;z^k \\ \overline{e^z} = \overline{\sum_{k=0}^\infty \frac{1}{k!}\;z^k} = \sum_{k=0}^\infty \overline{\;\frac{1}{k!}\;}\;\overline{z^k} = \sum_{k=0}^\infty \frac{1}{k!}\;{\overline{z}\;}^k = e^{\overline{z}} $$

Answer 3

But really, if $f$ is nice (analytic) and $f(z) \in \mathbb{R}$ for $z\in \mathbb{R}$, then $f(\bar z) = \overline{f(z)}$.