Show that a function is trivial on the unit circle Let $f$ : $\mathbb{S^1}$ $\to$ $\mathbb{R}$ continuous such that $\forall$ $z,z'$ $\in$ $\mathbb{S^1}$, $f(zz')=f(z)+f(z')$ 
Let $K=\{{e^{i\theta},\frac{\theta}{2\pi} \in \mathbb{Q}}\}$, show that f is trivial on $K$ i.e, $f(z)=0$  $\forall z \in K$    
Also what topological property $K$ enjoys inside  $\mathbb{S^1}$ $?$ I dont know how to start the proof and i need some hints please !!
 A: Let $z=1$, then $f(1z')=f(1)+f(z')$ and so $f(1)=0$. So, you know at least one element is always mapped to $0$. That's a good starting point for you I think.
Next, try looking at the case that $z=z'$. What about $z'=z^2$, $z'=z^n$?
A: $$\begin{cases}e^{\pm i\pi}=\cos(\pm\pi)+i\sin(\pm\pi)\pi=-1\in K\text{ because } \frac{\pm\pi}{2\pi}\in\Bbb Q\\e^{0i}=\cos 0+i\sin 0=1\in K \text{ because } \frac{0}{2\pi}\in\Bbb Q\end{cases}$$
Besides $f(\pm z)=f(\pm1)+f(z)\Rightarrow f(\pm1)=0$.
From $$\pi=\frac{\pi}{2}+\frac{\pi}{2}\\\frac{\pi}{2}=\frac{\pi}{4}+\frac{\pi}{4}\\\frac{\pi}{4}=\frac{\pi}{8}+\frac{\pi}{8}\\......\\......\\\frac{\pi}{2^n}=\frac{\pi}{2^{n+1}}+\frac{\pi}{2^{n+1}}$$ we get easily
$$f(e^{i\frac{\pi}{2}})= f(e^{i\frac{\pi}{4}})= f(e^{i\frac{\pi}{8}})=….= f(e^{i\frac{\pi}{2^n}})=0\text { for all } n$$
Furthermore $f(e^{i\frac{\pi}{2}})=f(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2})=f(i)=0$.
You can finish now with $f(z)=0$ thinking a little more.
On the other hand $K$ is easily seen to be a subgroup of the group multiplicative $\Bbb S^1$. Because of
$z\in K\iff rz\in K$ for all non-zero rational $r$ one can prove that $K$ is dense in $\Bbb S^1$. 
