Image of the upper half complex plane, under the function g(z) = exp(2 \pi i z).

[kalish1]

Problem: Given $W = \{z: z=x+iy, \ y>0\}$ and $g(z) = e^{2 \pi i z},$ what does the set $g(W)$ look like, and is it simply connected?

Attempt: $W$ represents the upper-half complex plane. And $$g(z) = e^{2 \pi i (x+iy)} = \cdots = e^{-2\pi y}(\cos (2 \pi x) + i \sin (2 \pi x)).$$ (Am I on the right track?)

I know simply connected means that there are no holes in the set, but I don't know how to describe the set geometrically.

Thanks in advance for help.

This question has been crossposted here: Image of the upper half complex plane, under the function $g(z) = e^{2\pi i z}$ - Mathematics Stack Exchange

 

[Euge]

Hi kalish,

Since $|\cos 2\pi x + i \sin 2\pi x| = 1$ for all $x \in \Bbb R$ and $0 < e^{-2\pi y} < 1$ for all $y > 0$, then $0 < |g(z)| < 1$ for all $z \in W$. Thus, $g$ maps $W$ into the punctured disc $\Bbb D^* = \{z \in \Bbb C : 0 < |z| < 1\}$. On the other hand, if $c \in \Bbb D^*$, then $|c| = |c|e^{2\pi i x}$ for some $x \in \Bbb R$, and $|c| = e^{-2\pi y}$ with $y = -(\log |c|)/(2\pi)$. Since $|c| < 1$, then $-\log |c| > 0$ and thus $y > 0$. Therefore, $c = g(x + yi)$ with $y > 0$. Consequently, $g$ maps $W$ onto $\Bbb D^*$.

The punctured disc $\Bbb D^*$ is not simply connected, for if $\gamma$ is the circle of radius $1/2$ centered at the origin (so it lies in $\Bbb D^*$), then $\int_\gamma \frac{dz}{z}$ is $2\pi i$, which is nonzero.