Suppose $f$ is an entire function such that $f(z_1+z_2)=f(z_1)+f(z_2)$ for all $z_1,z_2 \in \mathbb{C}$. Show that for all $z \in \mathbb{C}$, $f(z)=zf(1)$.
Here, by an entire function, we mean a function $f$ that is holomorphic on the whole of $\mathbb{C}$.
What I have so far is:
Since $f$ is differentiable everywhere, then for any $z \in \mathbb{C}$, we have $f'(z)=\mathop {\lim }\limits_{h \to 0} \frac{{f(z + h) - f(z)}}{h}$. But $f(z + h)=f(z)+f(h)$, so that $f'(z)=\mathop {\lim }\limits_{h \to 0} \frac{{f(h)}}{h}$. I'm not sure how to proceed from here. I saw somewhere that you should use $f(0)=0$, to get $f'(z)=f'(0)$, but how is $f(0)=0$?
Note that$$f(0)=f(0+0)=f(0)+f(0)$$and that therefore $f(0)=0$. So, for each $z\in\Bbb C$,$$f'(z)=\lim_{h\to0}\frac{f(h)}h=\lim_{h\to0}\frac{f(h)-f(0)}h=f'(0).$$Therefore, for some constant $k$, $(\forall z\in\Bbb C):f(z)=f'(0)z+k$. But $k=f(0)=0$. And then $f(1)=f'(0)\times1=f'(0)$. So, indeed, we always have $f(z)=zf(1)$.