Real part of a function

Question

Suppose we have the complex number $c$ and a rationale function $F(c)$. Could I say that $F(\mathrm{Re}(c))=\mathrm{Re}(F(c))$. Is there a proof of this?

Thank you!

Answer 1

No if the functions are allowed to have complex valued coefficients, e.g. $F(z)=i$

$$\DeclareMathOperator{\Re}{Re}\Re(F(z))=0\not=i=F(\Re(z)).$$

No if the function can have complex valued arguments, e.g. $F(z)=z^2$

$$\Re(F(i))=-1\not=0=F(\Re(i)).$$

Yes if the functions have real valued coefficients only and only real valued arguments are considered. For a given real number $x$ in the domain of $F$, we have $F(x)\in\Bbb R$ as well. Therefore

$$F(\Re(x))=\underbrace{F(x)}_{\in \,\Bbb R}=\Re(F(x)).$$