All solutions of $f(w+z)=f(w)f(z)$, $f(1)=e$ Let $w=u+iv$ and $z=x+iy$ with $u,v,x,y\in\mathbb{R}$. Is the function
$$f:\, \mathbb{C}\to\mathbb{C},\, f(z)=e^x\cos y+ie^x\sin y$$
the only solution of the functional equation
$$f:\, \mathbb{C}\to\mathbb{C},\, f(w+z)=f(w)f(z),\, f(1)=e?$$
I was only able to prove that it is a solution:
$f(1)=e$ trivially and
$$\begin{align}f(w)f(z)&=(e^u\cos v+ie^u\sin v)(e^x\cos y+ie^x\sin y)\\&=e^{u+x}(\cos v\cos y-\sin v\sin y)+ie^{u+x}(\sin v\cos y+\cos v\sin y)\\&=e^{u+x}\cos (v+y)+ie^{u+x}\sin (v+y)\\&=f(w+z).\end{align}$$
 A: Without any additional assumption the answer is no, it's not the only solution. Consider for example $f(z)=e^{\overline{z}}$, or in your terms $f(z)=e^x\cos y-ie^x\sin y$. Another example could be $f(z)=e^{\Re(z)}$ (or $f(z)=e^x$), I guess you get the idea.
If you add an assumption for $f$ to be holomorphic (for example) then the solution is unique, see Find the holomorphic function satisfying an equation and agrees with $e^x$ in real line.
A: THIS IS AN INCOMPLETE ANSWER:
Let's consider a couple of basic properties. First, $f(0) = f(0 + 0) = f(0)f(0)$, so either $f(0) = 0$ or $f(0) = 1$. But since $f(z) = f(z + 0)$ for any $z \in \mathbb{C}$, the option $f(0) = 0$ would lead to $f \equiv 0$ everywhere on $\mathbb{C}$. So let's assume $f(0) = 1$.
Working on that assumption, we have $1 = f(0) = f(z + (-z)) = f(z) f(-z)$, so $f(-z) = 1/f(z)$ when $f(z) \neq 0$.
Next, for any $0 < n \in \mathbb{Z}$, note that $$f(n) = f(1 + 1 + \dotsb + 1) = f(1) \dotsm f(1) = e^n$$ and, as a consequence, $f(-n) = e^{-n}$. Moreover, $$e = f(1) =  f(1/n + \dotsb + 1/n) = f(1/n) \dotsm f(1/n) = f(1/n)^{n},$$ so that $f(1/n) = e^{1/n}$.
For any rational $q = a/b$ with $a, b \in \mathbb{Z}$, $b \neq 0$, we then have $$f(q) = f(a/b) = f(1/b + 1/b + \dotsb + 1/b) = f(1/b) \dotsm f(1/b) = e^{1/b} \dotsm e^{1/b} = e^{a/b} = e^q.$$
If we want our function to be continuous, which seems a reasonable requirement to me, we need to have $$\lim_{n \to \infty} f(z_n) = f(\lim_{n \to \infty} z_n)$$ for all converging sequences $\{z_n\}_{n = 1, 2, \dotsc} \subset \mathbb{C}$. Considering this requirement for sequences of rationals, we have $$f(r) = f(\lim_{n \to \infty} q_n) =  \lim_{n \to \infty} f(q_n) = \lim_{n \to \infty} e^{q_n} = e^r$$ for $q_n \to r \in \mathbb{R}$. Since the rationals are dense in the reals, this is valid for all $r \in \mathbb{R}$, and $f$ restricted to $\mathbb R$ is indeed the exponential function.
To extend this to $\mathbb C$, we note that $f(x + iy) = f(x)f(iy) = e^x f(iy)$ for any $x, y \in \mathbb R$. Following similar considerations as the ones for the real case (e.g. $f(in) = f(i) \dotsm f(i) = f(i)^n$), we also conclude that $f(iy) = f(i)^y$ (note that in the real case we had $f(1 \cdot x) = f(1)^x = e^x$). Hence $$f(x + iy) = e^x f(i)^y.$$
There may be some other tricks you can pull off to determine $f(i)$, but I suspect you would need to introduce other conditions (e.g. differentiability) to fix it to one particular value. To conclude, the answer to your question is the following: up to the value of $f(i)$, there is only one non-trivial continuous function $f : \mathbb{C} \to \mathbb{C}$ satisfying $f(z + w) = f(z) f(w)$ for all $z, w \in \mathbb{C}$ and $f(1) = e$, namely $f(z) = e^{\Re(z)} f(i)^{\Im(z)}$.
