Simple identity for entire function

Question

If $f(-\overline{z})=-f(\overline{z})$ for all $z\in\mathbb{C}$, and if $f$ is entire, then $f(-z)=-f(z)$ for all $z\in\mathbb{C}$.

Is it false when the condition that $f$ is entire is removed from the above statement?

Answer 1

This has nothing to do with "entire " ! Take $w \in \mathbb C$ and let $z :=\overline{w}$. Then:

$$f(-z)=f(-\overline{w})=-f(\overline{w})=-f(z).$$

Answer 2

$f(-z)=-f(z)$ means that the graph of $f$ (for instance in this sense) looks the same if you turn it around $180^\circ$ about the origin (it has a $180^\circ$ rotational symmetry about the origin). $f(-\overline{z})=-f(\overline{z})$ means that if you mirror the graph of $f$ over the real axis, then the resulting graph has a $180^\circ$ rotational symmetry about the origin. It's quite clear intuitively that these two properties are equivalent.