Where does exponential function $f=e^z$ map lower half plane?

Question

Given exponential function $w=e^z$ , map the lower half plane. Can someone please help me what is the image of this function? I know that I have to write $z=x+iy$ , so for $w=u+iv$ and $w=e^x(cosy+isiny)$ , I got that $u=e^xcosy$ and $v=e^xsiny$ . In the lower half plane $x\in[-\infty,+\infty]$ and $y<0$ , but I', not sure what the image of this function will be(which set)... Any help is appreciated.

Answer 1

Let $w =u+iv\in \mathbb \setminus \{0\} $. Set $r = \sqrt{u^2+v^2}$ and set $x = \log r$. The latter is permitted because $w \neq 0$. Take $y$ to be the unique angle, $-\pi < y \leqslant \pi$ such that $ u = r\cos y $ and $ v = r\sin y$. Set $z = x+iy$. Now choose any $n \in \mathbb Z$, and then

$$\begin{align} e^{z + 2\pi i n} &= e^x e^{iy + 2\pi i n} \\ &= r e^{iy} \\&= r(\cos y + i\sin y) \\ &= u+iv \\ &= w \end{align}$$

So every $w$ is the image of a $z$. But $n$ is arbitrary, so taking $n$ large and negative, every $w$, except $0$, lies in the image of $e^z$ with $z$ in the lower half plane.