<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

Determine the region of the $w$ plane into which each of the
following is mapped by the transformation $w=z^2$.

\begin{description}
\item[(i)]
The first quadrant of the $z$-plane.

\item[(ii)]
The region bounded by $x=1,\ y=1,\ x+y=1$ (Hint: calculate the
real and imaginary coordinates of the transformed line segments
and sketch curves in the $w$-plane which they parametrise.)
\end{description}

\medskip

{\bf Answer}

Let $w=z^2,\ z=re^{i\theta}$. Then $w=r^2e^{2i\theta}$. Thus
points at $(r,\theta)$ are rotated by a further angle $\theta$ and
their modulus stretched by a factor $r$.

\begin{description}
\item[(i)]
First quadrant of $z$-plane

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$\stackrel{w=f(z)}{\longrightarrow}$ \setlength{\unitlength}{.5in}
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\put(0,0){\vector(1,0){4}}

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\put(4,0){\makebox(0,0)[l]{$u$}}
\end{picture}

\vspace{2in}

All points in 1st quadrant occupy $r&gt;0,\ 0 \leq \theta \leq
\ds\frac{\pi}{2}$. Thus all points in $w (=\rho e^{i\phi})$-plane
occupy $\rho &gt;0,\ 0\leq \phi \leq \pi$, i.e., the upper half of
the $w$-plane.

\newpage
\item[(ii)]
Region bounded by $x=1,\ y=1m\ x+y=1$.

If $w=z^2$ we have

$$(u+iv)=(x+iy)(x+iy)=x^2-y^2+2ixy$$

Therefore $\left\{\begin{array} {rcl} u &amp; = &amp; x^2-y^2\\ v &amp; = &amp;
2xy \end{array} \right.$

Thus the lines

$x=1 \rightarrow \left\{\begin{array}{rcl} u &amp; = &amp; 1-y^2\\ v &amp; = &amp;
2y \end{array}\right\} \Rightarrow u=1-\ds\frac{v^2}{4}$

$y=1 \rightarrow \left\{\begin{array}{rcl} u &amp; = &amp; x^2-1\\ v &amp; = &amp;
2x \end{array}\right\} \Rightarrow u=\ds\frac{v^2}{4}-1$

$\begin{array} {rcl} x+y &amp; = &amp; 1\\\rm{or}\ y &amp; = &amp; 1-x
\end{array}\rightarrow \left\{\begin{array}{rcl} u &amp; = &amp; x^2-(1-x)^2\\ &amp; =
&amp; 2x-1\\ v &amp; = &amp; 2x(1-x)\\ &amp; = &amp; 2x-2x^2
\end{array}\right\} \Rightarrow v=\ds\frac{1}{2}(1-u^2)$

Thus we have:

(z)

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\put(5,0){\makebox(0,0)[l]{$x$}}

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\put(2.5,.5){\makebox(0,0)[tr]{$x+y=1$}}
\end{picture}
$\stackrel{w=z^2}{\longrightarrow}$

PICTURE \vspace{2in}

Pick a points inside the region $ABC$ to see where it goes and
confirm $ABC \rightarrow A'B'C'$ shaded region.

Note that $w'=f(z)=2z$, so unless $z=0$ the local angles should be
conserved.

Thus is $\angle BAC=\angle ABC=\ds\frac{\pi}{4}$

then $\angle B'A'C'$ (or the tangents' angle at $A'$) is
$\ds\frac{\pi}{4}$

\ \ \ \ $\angle A'B'C'$ (or the tangents' angle at $B'$) is
$\ds\frac{\pi}{4}$

Similarly $\angle ACB=\ds\frac{\pi}{2}$ and $\angle A'C'B'$ is
$\ds\frac{\pi}{2}$
\end{description}

\end{document}
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