Every Lie group homomorphism from $U(1)$ to $\mathbb{R}$ is trivial I have to show that any Lie group homomorphism $f:U(1)\to (\mathbb{R},+)$ is the constant zero-map.
What I tried so far is the following: Let $z\in U(1)$. Then we have that
$$0=f(1)=f(\vert z\vert^{2})=f(z\cdot \overline{z})=f(z)+f(\overline{z})$$
However, this does not help me, since $\overline{z}=1/z$ and hence $f(\overline{z})=-f(z)$ by the properties of a group homomorphism. In other words, the equation above is always fulfilled. Any ideas?

Recall that homomorphisms between Lie groups are actually continuous. To me there are two ways to exploit the continuity to show the claim. I'll sketch the proof, in case you have troubles filling the gaps, let me know.
For the first version we note that any element of finite order in $U(1)$ gets mapped to zero (because $0=f(1)=f(z^n)=n f(z)$ implies $f(z)=0$ in $\mathbb{R}$). Show that the set of these elements is dense in $U(1)$ (consider for example $\{\exp(2\pi i q) \ : \ q \in \mathbb{Q}\}$) and then recall that $f$ is continuous to conclude.
Another way to see it is to prove first that $U(1)$ is compact (it is just the circle) and hence the image is compact and therefore bounded. Assume now you have $z\in U(1)$ such that $f(z)\neq 0$, then the orbit $\{ nf(z) \ : \ n\in \mathbb{N}\}=\{f(zn) \ : \ n\in \mathbb{N}\} \subseteq f(U(1))$ is unbounded, which gives the desired contradiction.
