I've got a problem to do something. Actually, we consider :
$U = \{ e^{s+it} \in \mathbb{C} \; | \; s,t \in \mathbb{R}, \; |s-t| < 1\}$,
We know that $U$ is an open and connected space, but first of all, I don't succeed to determine if $U$ is simply connected or not. Actually, I don't think so, but I don't know how to prove it. I thought about winding number, I would like to find two loop with different winding number, but I don't find out those loops...
Then, I have to show that it exists a complex logarithm on $U$ which verify that its value for $1$ is $2\pi i$, but I don't succeed to find which branch I have to cut... I have to play with the fact that $|s-t| < 1$ but how... ?
Thank you !
Edit : Actually, I don't have to cut a branch I think, because if I consider $e^{i\pi x}$ for $x \in \mathbb{R}$, I have $e^{\pi x+i\pi x} = e^{\pi x}e^{i\pi x} \in U$, so I can't just cut a branch, I have to do another way...
Actually, for $z = e^{s+it}$, I can define : $f(z) = s+i(t+2\pi)$. $f$ is well-defined car if $e^{s+it}=e^{s'+it'}$, then $s=s'$ and $t = t'$ mod $2\pi$, so as we have $|s-t| < 1$, we must have $t=t'$. So $f$ is well-defined, and we can verify that $e^{f(z)}=z$, so it's okay, no ?