Let $\mathscr A$ be an unital $C^*$-algebra and let $u \in \mathscr A$ be an unitary element (I.e., $u^*u=uu^*=1$). Is it true that
$$u=e^{ia}$$
for some hermitian element $a\in \mathscr A$?
Im not sure how to construct a counter example. I also think one should use functional calculus(?), but again, not sure how.
Here is a counterexample.
Consider $\mathcal A:=\mathcal C(\mathbb T)$, the $C^*$-algebra of all continuous (complex-valued) functions on the unit circle $\mathbb T:=\{ z\in\mathbb C;\; \vert z\vert=1\}$.
The function $u\in\mathcal C(\mathbb T)$ defined by $u(z):=z$ is certainly unitary. However, it is quite well known that one cannot define a continuous logarithm function on $\mathbb T$. So there is no continuous function $w$ on $\mathbb T$ such that $u=e^w$, and hence one cannot write $u=e^{ia}$ for some hermitian $a\in\mathcal A$.