### Author Topic: Quiz 3 LEC0101-1D  (Read 881 times)

#### yuxuan li

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##### Quiz 3 LEC0101-1D
« on: October 09, 2020, 01:15:54 PM »

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}