If $f(z)$ is entire and satisfying the condition $f(z_1+ z_2)=f(z_1) f(z_2) \exists c\in C$ such that $f(z)=e^{cz}$
I started complex analysis very recently How to prove this? I am not able to get any clue.
I thought in following lines
Take $g(z)=e^{-cz} f(z)$
$g'(z)=-ce^{-cz}f(z)+e^{-cz} f'(z)$
Again $f'(z) = \lim_{h \to 0} \frac{f(z+h)-f(z)}{h}$
I am not able to get $f(z_1 +z_2) = f(z_1)f(z_2)$ to explore further...
I assume $f(z+z')+f(z)f(z')$. Remark that $f(0.0)=f(0)^2$ implies that $f(0)=1$ or $f(0)=0$, if $f(0)=0, f(0+z)=f(0)f(z)=0$ and $f=0$. Suppose that $f\neq 0$. Write $c=f'(z)$
$f'(z)=lim_{h\rightarrow 0}{{f(z+h)-f(z)}\over h}=f(z)lim_{h\rightarrow 0}{{f(h)-1}\over h}=cf(z)$. consider $g(z)=e^{-cz}f(z)$, show that $g'(z)=-ce^{-z}f(z)+e^{-z}cf(z)=0$. This implies that $g$ is constant and $g(z)=g(0)=1$. We deduce that $f(z)=e^{cz}$.