Q:

Holomorphic function in complex analysis

For a holomorphic function $f$ we know that:

$f(x) \in R$, for every real number $x$

$f(1+i)=-1-i$

Determine $f(1-i).$

I'm really not seeing how this exercise can be solved. My best guess is the answer would be $f(1-i)=-1+i, $ but only because since the function sends real numbers in real numbers and $f(i+1)$ is its symmetric, then if $f(z)=-z$, it satisfies both these conditions.
But how do I prove this correctly? Is this connected in any way to the identity theorem?

A:

Let $g(z)=\overline{f\left(\overline z\right)}$. Then $g$ is holomorphic too and, if $x\in\Bbb R$, $g(x)=f(x)$. So, by the identity theorem, $g=f$. But then$$f(1-i)=f\left(\overline{1+i}\right)=\overline{f(1+i)}=\overline{-1-i}=-1+i.$$

