Find all homomorphisms from $\mathbb{C} \to \mathbb{C}$ Let $f(z):\mathbb{C} \to \mathbb{C}$ defined by $$z \mapsto f(z)$$
where $f(z)=a_0+a_1z+a_2z^2+\dots \in \mathbb{C}[[z]]$
My question is how to find all $f(z)\in \mathbb{C}[[z]]$ such that it is a homomorphism? That is for all $x,y \in \mathbb{C}$
$$f(x+y)=f(x)f(y)$$
What I have done is write down what $f(x+y)$ and $f(x)f(y)$ are and try to compare them term by term:
$$f(x+y)=a_0+a_1(x+y)+a_2(x+y)^2+\dots$$
$$f(x)f(y)=(a_0+a_1x+a_2x^2+\dots)\times(a_0+a_1y+a_2y^2+\dots)\\
=a_0^2+a_0a_1y+a_0a_2y^2+\dots\\+a_0a_1x+a_1^2xy+a_1a_2xy^2+\dots\\+a_0a_2x^2+a_1a_2x^2y+a_2^2x^2y^2+\dots\\\vdots$$
So $$a_0=a_0^2 \Rightarrow a_0=1 \\ 2a_2=a_1^2 \Rightarrow a_1=\pm \sqrt{2a_2}$$
It seems like all $a_i$ for $i\in \mathbb{N}_{>0}$ are linked together, meaning that if I set $a_1$ to be $0$ than all $a_i=0$. Hence $f(z)=1$ is a homomorphism, but than how can I find the rest?
 A: The series must, of course, be convergent for all $z$, so it defines an entire function. The homomorphism condition says
$$
f(w+z)=f(w)f(z)
$$
and differentiating with respect to $z$ gives
$$
f'(w+z)=f(w)f'(z)
$$
Therefore, with $z=0$, $f'(w)=f(w)f'(0)$. Note that $f(z)=0$ for some $z$ implies $f$ is constant: $f(w)=f((w-z)+z)=f(w-z)f(z)=0$. So we can well assume $f(z)\ne0$ for every $z$ and thus we can write
$$
f(z)=\exp(g(z))
$$
for some entire function $g$. Therefore
$$
g'(w)\exp(g(w))=\exp(g(z))f'(0)
$$
and so
$$
g'(w)=f'(0)
$$
and so $g(w)=a+f'(0)w$. Now, $f(0)=f(0+0)=f(0)f(0)$, so $f(0)=1$:
$$
1=\exp(a+f'(0)0)=\exp(a)
$$
and thus $a=2ki\pi$ for some integer $k$, which means
$$
f(z)=\exp(2ki\pi+f'(0)z)=\exp(f'(0)z)
$$
and $f'(0)$ can actually be any constant.
A: Let's assume for a second that you're really devoted to your method. As you note, we equate the coefficient of $x^iy^j$ on each side of
$$f(x+y)=f(x)f(y)$$
As you note, we have $a_0^2=a_0$ which implies $a_0=1$, assuming $0$ is excluded. Then, we just need to examine the coefficient of $xy^n$ on each side to find the only possible coefficients. In particular, the coefficient of $xy^n$ on the right is $a_1a_n$ and the coefficient on the left is, by the binomial theorem, $(n+1)a_{n+1}$. So, we have
$$a_{n+1}=\frac{a_1a_n}{n+1}.$$
By induction, we can write:
$$a_n=\frac{a_1^n}{n!}$$
noting that this holds for $n=0$ too. Then, we have that any such $f$ must have the form, where $\alpha=a_1$
$$f(x)=1+\alpha x + \frac{(\alpha x)^2}{2!}+\frac{(\alpha x)^3}{3!} + \frac{(\alpha x)^4}{4!}+\ldots$$
or, equivalently,
$$f(x)=e^{\alpha x}$$
for some $\alpha.$ you can verify that any choice of $\alpha$ yields a homomorphism.
