\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\ol}{\overline}
\parindent=0pt
\begin{document}

\begin{center}
{\bf Complex Numbers}
\end{center}

{\bf Geometrical Transformations in the Complex Plane}

For functions of a real variable such as $f(x)=\sin x, \,\,
g(x)=x^2+2$ etc you are used to illustrating these geometrically,
usually on a cartesian graph.  If we have functions of a complex
variable given by equations such as $w=\sin z$ or $w=z^2+2$ we
cannot use a cartesian graph, since $z$ cannot be represented on
an ordered axis.  Indeed $z$ may range over the whole of the two
dimensional complex plane, so that if $w$ is also complex we would
need a 4-dimensional space to plot a graph such as $w=z^2+2$. Most
of us cannot visualise this, and what we usually do is to have two
copies of the complex plane, and we look at points in the
$z$-plane and see how they are transformed into points in the
$w$-plane.  We also look at sets of points, curves or regions in
the $z$-plane and their images in the $w$-plane.

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Examples

\begin{itemize}
\item[1)]
$w=f(z)=z+2$.  This simply shifts every point two units in the
direction of the real axis - it is a translation.

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\put(1.3,0){\makebox(0,0){$z-plane$}}


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\put(8.2,1.7){\makebox(0,0){$B'$}}

\put(6.5,0){\makebox(0,0){$w-plane$}}

\put(3.5,1.2){\makebox(0,0){$z\rightarrow z+2$}}
\end{picture}

\newpage

\item[2)]
$w=z+2-i$, again a translation

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\put(1.3,1){\makebox(0,0){$z-plane$}}


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\put(5,0){\line(0,1){3}}

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\put(8,0.5){\line(0,1){1}}

\put(7,0.5){\line(0,1){1}}

\put(6.7,0.7){\makebox(0,0){$D'$}}

\put(6.7,1.7){\makebox(0,0){$A'$}}

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\put(8.2,1.7){\makebox(0,0){$B'$}}

\put(6.5,0){\makebox(0,0){$w-plane$}}

\put(3.5,2.2){\makebox(0,0){$z\rightarrow z+2-i$}}
\end{picture}

\item[3)]
$w=\ol{z}+2$, this is not a translation.

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\put(1.3,1){\makebox(0,0){$z-plane$}}


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\put(6.5,0){\makebox(0,0){$w-plane$}}

\put(3.5,2.2){\makebox(0,0){$z\rightarrow \ol{z}+2$}}
\end{picture}


\item[4)]
$w=2z$  Now $|w|=2|z| \,\,\, \arg w=\arg2+\arg z=\arg z$

So this is an enlargement about the origin with scale factor 2.

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\put(6.5,0){\makebox(0,0){$w-plane$}}

\put(3.5,1.2){\makebox(0,0){$z\rightarrow 2z$}}
\end{picture}

\item[5)]
$\ds w=iz \,\,\, |w|=|z| \,\,\, \arg w=\arg i+\arg
z=\frac{\pi}{2}+\arg z$

So this is a rotation through $\frac{\pi}{2}$ anticlockwise about
$O$.

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\put(5.5,0){\makebox(0,0){$w-plane$}}

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\end{itemize}


In general if $\alpha$ is any complex number and we write
$\alpha=re^{i\theta}$ then $w=\alpha z$ is an enlargement by scale
factor $r$ together with a rotation about $O$ through the angle
$\theta$ anticlockwise.

If we write \begin{eqnarray*} \alpha &=& a+ib\\ z &=& x+iy\\ w &=&
u+iv\end{eqnarray*}

then $w=\alpha z$

becomes $u+iv=(a+ib)(x+iy)$

and so \begin{eqnarray*} u &=& ax-by\\ v &=& bx+ay \end{eqnarray*}

We write this in the form $\left(\begin{array}{c} u\\ v\end{array}
\right)=\left(\begin{array}{cc} a & -b\\ b & a\end{array}\right)
\left(\begin{array}{c} x\\ y\end{array}\right)$

The right hand side can be interpreted as a multiplication, but at
the moment it seems a rather odd kind of multiplication.

We call $\left(\begin{array}{c} x\\ y\end{array}\right)$ a column
vector.

We call $\left(\begin{array}{cc} a & -b\\ b & a\end{array}\right)$
a matrix.

If we now have another transformation $\xi=\beta w$ where
$\beta=c+id$ then if we write $\xi=s+it$ we shall have

\begin{eqnarray*} \left(\begin{array}{c} s\\ t\end{array}\right)&=&\left(
\begin{array}{cc} c & -d\\ d & c\end{array}\right)
\left(\begin{array}{c} u\\ v\end{array}\right)\\ &=& \left(
\begin{array}{cc} c & -d\\ d & c\end{array}\right)\left(
\begin{array}{cc} a & -b\\ b & a\end{array}\right)
\left(\begin{array}{c} x\\ y\end{array}\right)\end{eqnarray*}

If we now do the substitutions

$\hspace{0.5in} s=cu-dv$

$\hspace{0.5in} t=du+cv$

in the first pair of equations we get

$\hspace{0.5in} s=(ca-db)x-(cb+da)y$

$\hspace{0.5in} t=(ad+bc)x+(ac-bd)y$

$\left(\begin{array}{c} s\\t\end{array}\right) =
\left(\begin{array}{cc} (ca-bd) & -(cb+da)\\ (ad+bc) & (ac-bd)
\end{array}\right) \left(\begin{array}{c} x\\ y\end{array}\right)$

This suggests that we should define

$\left(\begin{array}{cc} c & -d\\ d & c\end{array}\right)\left(
\begin{array}{cc} a & -b\\ b & a\end{array}\right)=\left(\begin{array}{cc}
(ca-bd) & -(cb+da)\\ (ad+bc) & (ac-bd) \end{array}\right)$

Finally if we go back to the original equation $w=\alpha z
\,\,\,\, v=\beta w$ we obtain $\xi=\beta\alpha z$ and
$\beta\alpha=(c+id)(a+ib)=(ac-bd)+i(ad+bc)$

If we write $\alpha$ and $\beta$ in polar form, taking $r=1$ for
both, so that they both correspond to rotations, we then have

$\hspace{0.5in} \alpha =\cos\theta+i\sin\theta$

$\hspace{0.5in} \beta =\cos\phi+i\sin\phi$

The corresponding matrices are

$\left(\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta &
\cos\theta\end{array}\right) \left(\begin{array}{cc} \cos\phi &
-\sin\phi\\ \sin\phi & \cos\phi\end{array}\right)$

$=\left(\begin{array}{cc} \cos\theta\cos\phi-\sin\theta\sin\phi &
-(\cos\theta\sin\phi+\sin\theta\cos\phi)\\
\sin\theta\cos\phi+\cos\theta\sin\phi &
\cos\theta\cos\phi-\sin\theta\sin\phi \end{array}\right)$

$=\left(\begin{array}{cc} \cos(\theta+\phi) & -\sin(\theta+\phi)\\
\sin(\theta+\phi) & \cos(\theta+\phi)\end{array}\right)$

which is in accordance with what we found previously.

${}$

Notice that although each complex number can be represented by a
matrix, matrices such as $\left(\begin{array}{cc} 1&1\\
0&1\end{array}\right)$ do not correspond to complex numbers.  We
can nevertheless use them to transform the plane.

$\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)
\left(\begin{array}{c}x\\ y\end{array}\right) =
\left(\begin{array}{c}x+y\\ y\end{array}\right)$

This corresponds to a shearing transformation.

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\put(4.5,2){\makebox(0,0){$\longrightarrow$}}


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In considering matrices used as transformations we have so far
considered the problem of finding the image of given points.

$A\left(\begin{array}{c}x\\
y\end{array}\right)=\left(\begin{array}{c}X\\ Y\end{array}\right)$

i.e. given $\left(\begin{array}{c}x\\ y\end{array}\right)$ what is
$\left(\begin{array}{c}X\\ Y\end{array}\right)$?

We now consider the reverse problem:

given $\left(\begin{array}{c}X\\ Y\end{array}\right)$ what is
$\left(\begin{array}{c}x\\ y\end{array}\right)$?

$\left(\begin{array}{cc}a&b\\ c&d\end{array}\right)
\left(\begin{array}{c}x\\ y\end{array}\right)=
\left(\begin{array}{c}X\\ Y\end{array}\right)$

so

$\begin{array}{rclc}ax+by&=&X& (1)\\cx+dy&=&Y& (2)\end{array}$

$(1)*d$ and $(2)*b \Rightarrow$

$\begin{array}{rclc}adx+bdy&=&dX\\bcx+bdy&=&bY\end{array}$

subtracting gives

$\begin{array}{rclc}(ad-bc)x&=&dX-bY& (3)\end{array}$

$(1)*c$ and $(2)*a \Rightarrow$

$\begin{array}{rclc}acx+bcy&=&cX\\acx+ady&=&aY\end{array}$

subtracting gives

$\begin{array}{rclc}(ad-bc)y&=&aY-cX& (4)\end{array}$

(3) and (4) can be solved for $x$ and $y$ iff $ad-bc\not=0$.  If
$ad-bc\not=0$ we then have

$\ds x=\frac{d}{ad-bc}X-\frac{b}{ad-bc}Y$

$\ds y=\frac{-c}{ad-bc}X+\frac{a}{ad-bc}Y$

so

$\begin{array}{ccl}\left(\begin{array}{c}x\\
y\end{array}\right)&=&\left(\begin{array}{cc} \frac{d}{ad-bc}&
\frac{-b}{ad-bc}\\ \frac{-c}{ad-bc}&
\frac{a}{ad-bc}\end{array}\right) \left(\begin{array}{c}X\\
Y\end{array}\right)\\ &=&
\frac{1}{ad-bc}\left(\begin{array}{cc}d&-b\\
-c&a\end{array}\right)\left(\begin{array}{c}X\\
Y\end{array}\right)\\ &=& \frac{1}{\triangle}
\left(\begin{array}{cc}d&-b\\ -c&a\end{array}\right)\end{array}$

The matrix

$\left(\begin{array}{cc}\frac{d}{\triangle}&
\frac{-b}{\triangle}\\ \frac{-c}{\triangle}&
\frac{a}{\triangle}\end{array}\right)$

is called the inverse of $A=\left(\begin{array}{cc}a&b\\
c&d\end{array}\right)$ written $A^{-1}$

$A^{-1}A=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)$

As a transformation this matrix does nothing at all.  All points
are fixed.  It is called the identity matrix.

$\triangle=ad-bc$ is called the determinant of $A$. So $A$ has an
inverse iff its determinant is non-zero.

For a complex number matrix

$\alpha=\left(\begin{array}{cc}a&-b\\ b&a\end{array}\right)
\begin{array}{l}\triangle=a^2+b^2=|\alpha|^2\\ \triangle=0 {\rm \
iff\ } a=b=0 {\rm \ \ i.e.\ } \alpha=0\end{array}$

and its inverse is

$\ds \frac{1}{|\alpha|^2}\left(\begin{array}{cc}a&b\\
-b&a\end{array}\right)=\frac{\ol{\alpha}}{|\alpha|^2}
=\frac{1}{\alpha} \hspace{0.3in} \alpha\not=0$

In widening the system to include all possible $2\times2$ matrices
we have included many matrices which do not have inverses.  We
have also sacrificed commutativity of multiplication, as $AB$ does
not always equal $BA$.

However we can deal with many different transformations, and
matrices turn out to have many and varied applications.

{\bf Other transformations}

There are many transformations not represented by $2\times 2$
matrices as above.  As an example we consider a few properties of
the transformation $w=z^2$.  It is convenient to use polar
co-ordinates, we use $(r,\theta)$ in the $z$-plane and $(p,\phi)$
in the $w$-plane.

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\put(2.5,3){\makebox(0,0){$z$-plane}}


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$\hspace{1.5in}\begin{array}{ll} A(1,0)&A'(1,0)\\
B(\sqrt2,\frac{\pi}{4})&B'(2,\frac{\pi}{2}\\
C(1,\frac{pi}{2}&C'(1,\pi)\\ D(1,\pi)&D'(1,2\pi)=(1,0)\\
E(1,\frac{3\pi}{2}&E'(1,3\pi)=(1,\pi)\end{array}$


DIAGRAM


so $z=e^{i\theta} \,\,\,\, -\pi\leq\theta\leq\pi$ corresponds to a
circle traced twice in the $w$-plane.


DIAGRAM


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\put(8.5,0.5){\makebox(0,0){plane without +ve real axis}}

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Reverting to cartesians now let $z=x+iy \,\,\,\, w=\xi+i\eta$

$\xi+i\eta=x^2-y^2+2ixy$ so $\xi=x^2-y^2 \,\,\,\, \eta=2xy$

Now if $x=1, \,\,\,\, \xi=1-y^2 \,\,\,\, \eta=2y$

so $\ds\xi=1-\frac{\eta^2}{4}$


DIAGRAM


If $y=1 \,\,\,\, \xi=x^2-1 \,\,\,\, \eta=2x$ so
$\ds\xi=\frac{\eta^2}{4}-1$


DIAGRAMS


\end{document}