Homomorphism and Holomorphic Functions Let $M \leq \mathbb{C}$ be the additive subgroup such that $$M=\{n+im \in \mathbb{C}:n,m\in \mathbb{Z} \}.$$ Let $f$ be a entire function such that $f(0)=0$. Suppose that $$g(z+M)=f(z)+M, (z\in\mathbb{C})$$
defines a group homomorphism, $g:\mathbb{C}/M \rightarrow\mathbb{C}/M$, where $\mathbb{C}/M$ is the quotient group.
$(a)$ Prove that exists some disk $D$ centered on $0$ such that $f(z+w)=f(z)+f(w)$ for all $z,w \in D.$
$(b)$ Prove that exists some $b\in\mathbb{C}$ such that $f(z)=bz$ for all $z$ in some neighborhood of $0$.
$(c)$ Prove that exists some $b\in \mathbb{C}$, such that $g(z+M)=bz+M$ for all $z \in \mathbb{C}.$
I don't see a clear path, because we never use group properties in the complex variable course. However, as $f$ is entire and $f(0)=0$, there exists a holomorphic function $h$ such that $f(z)=z^rh(z)$ for some $r\geq1$, and $h(0)\neq0$, but how can I use this?
Even so is not clear to me how to use the quotient, and which properties of the Gaussian integers are important in this case.
Edit: I tried this for example. As $g$ is a homomorphism, then
$$g(z+M+w+M)=g(z+M)+g(w+M).$$
This implies that
$$f(z+w)=f(z)+f(w)+n+im$$
for some $n,m \in \mathbb{Z}$, but how do I "get rid" of $n+im$??

Notice that $f(z+w)-f(z)-f(w)$ is a continuous function in $z$ and $w$ and that $f(0+0)-f(0)-f(0)=0.$ We have $f(z+w)-f(z)-f(w)\in M$ for all $z,w\in \mathbb{C}$ and $M$ is a discrete set. Suppose that there exist some $z,w\in \mathbb{C}^2$ such that $f(z+w)-f(z)-f(w)\neq 0.$ Then $f^{-1}(0)\subseteq \mathbb{C}^2$ is both closed and open and $\emptyset\subsetneq f^{-1}(0)\subsetneq \mathbb{C}^2$, contradicting the fact that $\mathbb{C}^2$ is connected.
This answers part (a) of your question.
