$f(z)$ entire function satisfying $f(z_{1} + {z_{2}})=f(z_{1})+f(z_{2})$.

Question

Let $f(z)$ be an entire function satisfying $$f(z_{1} + z_{2})=f(z_{1})+f(z_{2}) $$ What is the form of f ?

$f$ is entire $\implies$ $f'(z)\ \forall z \implies f'(z)=f'(0)$ Now : $$f'(0)=\lim_{h\to 0}\frac{f(h+0)-f(0)}{h}=0$$ which implies that $$f(z)= c $$ Isn't $f(z)=az$ is the correct answer for this question ?

Answer 1

Taking derivatives with respect to $z_1$ gives with the chain rule that $$ f'(z_1 + z_2) = f'(z_1) $$ for all $z_1, z_2 \in \mathbf C$. Hence there is some $a \in \mathbf C$ such that $f'(z) = a$ for all $z \in \mathbf C$. Therefore $f(z) = az+b$ for some $b \in \mathbf C$. We have $$ f(z_1 + z_2) = az_1 + az_2 + b \stackrel!= az_1 + az_2 + 2b = f(z_1) + f(z_2) $$ This holds iff $b = 0$, hence $f(z) = az$ for all $z\in \mathbf C$.

Answer 2

We have $f(0)=f(0+0)=f(0)+f(0)$, hence $f(0)=0$,

For $z \in \mathbb C$:

$\frac{f(z+h)-f(z)}{h}=\frac{f(h)}{h}=\frac{f(h)-f(0)}{h} \to f'(0)$ for $h \to 0$,

hence $f'(z)=f'(0)$ for all $z$.

Thus $f'$ is constant.