Problem:
Show that $F(z) = e^{z}$ maps the strip $S = \{x+iy: -\infty < x < \infty, -\pi/2 \leq y \leq \pi/2\}$ onto the region $D = \{w = s + it: s \geq 0, w \ne 0\}$ and that $F$ is one-to-one on $S$. Furthermore, show that $F$ maps the boundary of $S$ onto all the boundary of $Q$ except $\{w = 0\}$. Explain what happens to each of the horizontal lines $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$.
Proof:
Firstly, we are going to expand the equation and prove that $F$ maps the strip $S$ onto the region $D$.
Let $F(z) = z_0 = u_0 + iv_0$.
$F(z) = e^{z} = e^{x+iy} = e^{x} \cdot e^{iy} = e^{x} \cdot (cos(y) + isin(y)) = e^{x}cos(y) + ie^{x}sin(y) = u_0 + iv_0$.
Note $x \in \mathbb{R}, y \in [-\pi/2, \pi/2]$, we may conclude that $e^{x} \ge 0, cos(y) \in [0,1], and \sin(y) \in [-1,1]$.
Then we have $Re(z_0) = u_0 = e^{x}cos(y) \ge 0$.
Although we get $Re(z_0) = u_0 = 0$ when $x = \pi/2$ or $-\pi/2, Im(z_0) \ne 0$. Thus after mapping, $z_0$ takes all positive x-axis except origin.
Moreover, we have $Im(z_0) = v_0 = e^{x}sin(y) \in \mathbb{R}$.
Since $z_0$ is an arbitrary element we chose from S, we have proved S gets mapped onto D.
Secondly, we are proving $F(z)$ is one-to-one mapping on S.
Equivalently, WTS $F(z_1) = F(z_2) \Longrightarrow z_1 = z_2$.
$\begin{gather}
F(z_1) = F(z_2) \\
e^{x_1}\cdot e^{iy_1} = e^{x_2} \cdot e^{iy_2} \\
{\frac{e^{x_1} \cdot e^{iy_1}}{e^{x_2}\cdot e^{iy_2}}} = 1 \\
e^{x_1-x_2} \cdot e^{i(y_1-y_2)} = 1
\end{gather}$
That implies $Re(z_1) = Re(z_2)$ and $Im(z_1) = Im(z_2)$, alternatively, $x_1 = x_2$ and $y_1 = y_2$
Therefore we have proved F is one-to-one on S.
Lastly, we will discuss what happens to the boundary $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$.
When $y = \frac {\pi}{2}$:
$\begin{gather}
\begin{split}
F(x + i\pi/2) & = e^{x} \cdot e^{\frac {i\pi}{2}} \\
& = e^{x} \cdot (cos(\pi/2) + isin(\pi/2)) \\
& = ie^{x}
\end{split}
\end{gather}$
When $y = \frac {-i\pi}{2}$:
$\begin{gather}
\begin{split}
F(x - i\pi/2) & = e^{x} \cdot e^{\frac {-i\pi}{2}} \\
& = e^{x} \cdot (cos(-\pi/2) + isin(-\pi/2)) \\
& = -ie^{x}
\end{split}
\end{gather}$
Since $e^{x}$ > 0, boundaries $\{Im z = \pi/2\}$ and $\{Im z = -\pi/2\}$ is mapped to positive y-axis and negative y-axis except the origin 0.
The pic below is the domain before and after mapping.
