For an entire function $f(z)$, is it true that $f(z)=f(\bar{z})$ for all complex $z$?

Question

I know that $f(z)$ is entire so I can write it as Taylor series: $f(z)=\sum c_n z^n=\sum Re(c_n)z^n+i\sum Im(c_n)z^n$

from above we see this:

$\overline{f(z)}=\sum c_n z^n=\sum Re(c_n)z^n-i\sum Im(c_n)z^n$

(Pay attention for the minus!)

and this:

$f(\bar{z})=\sum c_n \bar{z}^n=\sum Re(c_n)\bar{z}^n+i\sum Im(c_n)\bar{z}^n$

after playing with some algebra i found that $f(\bar{z})$ equals to $\overline{\overline{f(z)}}=f(z)$

it sound strange, someone can tell me if it is really true for entire $f(z)$ for every complex $z$ ?

EDIT

ok I see that it is no true, but what about $|f(\bar{z})|$ ? is it equals to absolute value of something? maybe $|\overline{f(z)}|$ ?

Answer 1

No, it is not true. Just take $f(z)=iz$, which is an entire function.

Note that$$f(z)=\sum_{n=0}^\infty c_nz^n\implies\overline{f(z)}=\sum_{n=0}^\infty\overline{c_n}\,\overline z^n.$$

Answer 2

Let $f(z)=z+i$. Then $f(\overline z)=\overline z+i$. There is no particular relation bewteen

$$|f(z)|^2=x^2+(y+1)^2$$ and $$|f(\overline z)|^2=x^2+(y-1).$$