Problem: Show that the function $w=g(z)=e^{z^{2}}$ maps the lines {$x = y$} and {$x = -y$} onto the circle {$|w|=1$}.
Show further that g maps each of the two pieces of the region {$x+iy: x^{2}>y^{2}$} onto the set {$w: |w|>1$}.
$
\begin{align}
\text{when }x^{2}=y^{2}\text{:}
\text{ when x = y: } \notag\\
\text{ }g(z)&=g(x+ix)\notag\\
&=e^{x^{2}+2ix^{2}-x^{2}}\notag\\
&=e^{2ix^{2}}\notag\\
&=\cos{(2x^{2})}+i\sin{(2x^{2})}\notag\\
&=x'+iy'\text{, set } x'=\cos{(2x^{2})}\text{ and }y'=\sin{(2x^{2})}\text{,}\notag\\
&\text{here r = 1, and }x'^{2}+y'^{2}=1\notag\\
\text{ when x = -y: } \notag\\
\text{ }g(z)&=g(x-ix)\notag\\
&=e^{x^{2}-2ix^{2}-x^{2}}\notag\\
&=e^{-2ix^{2}}\notag\\
&=\cos{(-2x^{2})}+i\sin{(-2x^{2})}\notag\\
&=x'+iy'\text{, set } x'=\cos{(-2x^{2})}\text{ and }y'=\sin{(-2x^{2})}\text{,}\notag\\
&\text{here r = 1, and }x'^{2}+y'^{2}=1\notag\\
&\text{Therefore, g(z) maps x = y and x = -y onto {w: |w| = 1}}\notag\\
\end{align}
$
$
\begin{align}
\text{ when }x^{2}>y^{2}\text{:}\notag\\
\text{ }g(z)&=g(x+iy)\notag\\
&=e^{{(x+iy)}^{2}}\notag\\
&=e^{x^{2}-y^{2}+2ixy}\notag\\
&=e^{x^{2}-y^{2}}e^{2ixy}\text{,}\notag\\
&\text{here }r=e^{x^{2}-y^{2}}\text{, which is greater than 1 when }x^{2}>y^{2}\notag\\
&\text{Therefore, g(z) maps it onto {w: |w| > 1}}\notag\\
\text{ when }x^{2}<y^{2}\text{:}\notag\\
\text{ }g(z)&=g(x+iy)\notag\\
&=e^{{(x+iy)}^{2}}\notag\\
&=e^{x^{2}-y^{2}+2ixy}\notag\\
&=e^{x^{2}-y^{2}}e^{2ixy}\text{,}\notag\\
&\text{here }r=e^{x^{2}-y^{2}}\text{, which is smaller than 1 when }x^{2}<y^{2}\notag\\
&\text{Therefore, g(z) maps it onto {w: |w| < 1}}\notag\\
\end{align}
$
