$f(z)= az$ 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 $f(z) = az$.
