$f(z)= az$ 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?
It's true under weaker assumptions, but let's do it by assuming that $f$ is analytic.

Fix $w \in \mathbb{C}$. Since $f(z+w) = f(z)+f(w)$, it follows that $f'(z+w) = f'(z)$ for all $z$. Hence $f'$ is constant, say $f'(z) = a$ which implies that $f(z) = az+c$.

Plug in $z_1 = z_2 = 0$ in the defining equation to conclude that $f(0) = 0$, so $f(z) = az$.
$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 $