$f(z)= az$ if $f$ is analytic and $f(z_{1}+z_{2})=f(z_{1})+f(z_{2})$ If $f$ is an analytic function with $f(z_{1}+z_{2})=f(z_{1})+f(z_{2})$, how can we show  that $f(z)= az$  where $a$  is a complex constant?

$f'(z)=\ lt_{h\to\ 0}\frac{f(z+h)-f(z)}{h}=lt_{h\to\ 0}\frac{f(z)+f(h)-f(z)}{h}\\=lt_{h\to\ 0}\frac{f(h)}{h}=lt_{h\to\ 0}\frac{f(0)+ \frac{hf'(0)}{1!}+\frac{h^2f''(0)}{2!}+...}{h} $
(by the taylor's series expansion  of $f(h)$
)
$so, f'(z)=lt_{h\to\ 0}\frac{f'(0)}{1!}+\frac{hf''(0)}{2!}+...$
since , $f'(0)$ is a complex constant,  then, $f'(z)$ is constant. 
Then 
$f(z) =az+b$ since $f(0)=0$, then we have that $f(z)=az $
