Functional equation $f(x+y)=f(x)f(y)$ for complex-valued $f$ It is well known that the only continuous functions $f: \mathbb{R} \to \mathbb{R}$ which satisfy the equation
$$f(x+y)=f(x)f(y)$$
are the exponential functions $f(x)=a^x$.
I am trying to prove a similar result when we allow $f$ to take complex values, i.e. if $f$ is a continuous function from $\mathbb{R}$ to $\mathbb{C}$ which satisfies $f(x+y)=f(x)f(y)$ for all $x,y \in \mathbb{R}$, then there exists some $z\in \mathbb{C}$ such that $f(x)=e^{zx}$. However, I am totally stuck after attempting a few different methods.
In real analysis, the steps towards proving this result is: (1) Use induction to show that $f(x)=a^x$ is true for all $x\in \mathbb{Z}$; (2) Use $f(1)=(f(\frac{1}{n}))^n$ to show that $f(\frac{1}{n})=(f(1))^{\frac{1}{n}}$, hence $f(x)=a^x$ is true for all $x\in \mathbb{Q}$; and (3) Use continuity to extend this to all $x\in \mathbb{R}$. However, when we allow $f$ to take value in $\mathbb{C}$, it seems that we're stuck at step (2), since every complex number has n distinct complex n-th roots and we do not know which one to choose for $f(\frac{1}{n})$. We do have the additional assumption of continuity, but I don't quite see how it can be employed in this step.
Any help will be appreciated.
 A: First we find all continuous maps $f:%
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%BeginExpansion
\mathbb{R}
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\rightarrow S^{1}$ such that $f(x+y)=f(x)f(y)$ for all $x,y.$ Then we replace  the range by $\mathbb C$.
Note that $f(0)=1$. Fix a positive integer $N$. By a standard
argument in Complex Analysis there exists a unique continuous function $%
h_{N}:[-N,N]\rightarrow 
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%BeginExpansion
\mathbb{R}
%EndExpansion
$ such that $f(x)=e^{ih_{N}(x)}$ $(\left\vert x\right\vert \leq N)$ and $%
h_{N}(0)=1$. It follows easily that $h_{N}^{\prime }s$ define a continuous
function $h:%
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\mathbb{R}
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\rightarrow 
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\mathbb{R}
%EndExpansion
$ such that $h(0)=0$ and $f(x)=e^{ih(x)}$ for all real numbers $x$. Note
that $e^{i[h(a+b)-h(a)-h(b)]}=1$ so $h(a+b)-h(a)-h(b)=2n\pi $ for some
integer $n$. By continuity of $h$ we conclude that $n$ does not depend on $a$
and $b$. Since $h(0)=0$ we conclude that $h$ is additive.
Since $h$ is additive and continuous there is a real number $a$ such that $%
h(x)=ax$ for all $x$. Hence $f(x)=e^{iax}$. Now consider the second case.
Since $f(0)=f^{2}(0)$ either $f(0)=0$ or $f(0)=1$. If $f(x)=0$ for some $x$
then $f(x+y)=f(x)f(y)=0$ for all $y$ which gives $f\equiv 0$. If this is not
the case then $f(0)=1$ and $f$ never vanishes. Let $g(x)=\frac{f(x)}{%
\left\vert f(x)\right\vert }$. The first part can be applied to $g$ and we
get $f(x)=e^{iax}\left\vert f(x)\right\vert $. Also $\log \left\vert
f(x)\right\vert $ is an additive continuous function on $%
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%BeginExpansion
\mathbb{R}
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,$ so $\left\vert f(x)\right\vert =e^{bx}$ for some real number $b$. We now
have $f(x)=e^{(b+ia)x}$.
