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{\bf Complex Numbers}
\end{center}

\textbf{History}

\lq\lq The historical development of complex number\rq\rq
D.R.Green Mathematical Gazette June 1976 pp99-107.

In {\bf N} we cannot solve $x+2=1$

In {\bf Z} we cannot solve $2x=1$

In {\bf Q} we cannot solve $x^2=2$

In {\bf R} we cannot solve $x^2+1=0$

You have all done some work on complex numbers, and this
introduction is in the spirit of the construction from {\bf Z} to
{\bf Q}.

\bigskip

\textbf{Definition}

A complex number is an ordered pair $(x,y)$ of real numbers, with
addition and multiplication defined by

\begin{eqnarray*} (x,y)+(x',y')&=&(x+x',y+y')\\
(x,y).(x',y')&=&(xx'-yy',xy'+yx')\end{eqnarray*}

With these definitions the complex number system {\bf C} has all
the properties of a field.  Now we have

\begin{eqnarray*} (x,0)+(x',0)&=&(x+x',0)\\
(x,0)(x',0)&=&(xx',0)\end{eqnarray*}

so there is a subsystem which behaves like {\bf R}.

$(0,1)(0,1)=(-1,0)$

$(x,y)=(x,0)+(0,y)=(x,0)+(y,0)(0,1)$

We shall abbreviate $(x,0)$ to $x$ and $(0,1)$ to $i$.

So we write $(x,y)=x+yi$

$x$ is called the \emph{real part} of the complex number.

$y$ is called the \emph{imaginary part} of the complex number.

Using this new symbolism we have:

$(x+yi)+(x'+y'i)=(x+x')+(y+y')i$

$(x+yi)(x'+y'i)=(xx'-yy')+(xy'+yx')i$



{\bf The Complex Plane}

We can represent $x+yi$ as a point in the plane with coordinates
$(x,y)$.

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If we write $z=x+yi$ then we have $x=r\cos\theta \hspace{0.3in}
y=r\sin\theta$

So $z=r(\cos\theta+i\sin\theta)$ - Polar form of $z$.

$r$ is called the modulus of $z$; $|z|=\sqrt{(x^2+y^2)}$

$\theta$ is called the argument of $z$; $\arg z$ it satisfies
$\ds\tan\theta=\frac{y}{x}$

There are many values of $\theta$ satisfying
$\tan\theta=\frac{y}{x}$.  The value of $\theta$ is taken to
satisfy $-\pi<\theta\leq\pi$ and this is called the principal
argument of $z$.

So

$\begin{array}{lcc} \arg(1+i) & = & \ds\frac{\pi}{4}\\ \arg i & =
& \ds\frac{\pi}{2}\\ \arg -1 & = & \pi\\ \arg(-1-i) & = &
\ds-\frac{3\pi}{4}\end{array}$

Note that to say $\ds\theta=\tan^{-1}\frac{y}{x}$ is not correct,
for it does not distinguish

$1+i \,\,\, (x=y=1)$ from $-1-i \,\,\, (x=y=-1)$.

Addition in the complex plane is interpreted geometrically through
the parallelogram law.

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Triangle inequality $|z+z'|\leq |z|+|z'|$

\bigskip

\textbf{Example}

Prove from the triangle inequality that

$||z|-|z'||\leq |z+z'|$

\bigskip

$|z|-|z'|=|(z+z')-z'|-|z'|\leq|z+z'|+|z'|-|z'|=|z+z'|$

Similarly $|z'|-|z|\leq|z'+z|$

Thus $||z|-|z'||\leq|z+z'|$

\bigskip

Multiplication is best approached using the polar form.

Let $z=r(\cos\theta+i\sin\theta); \hspace{0.5in}
z'=r'(\cos\theta'+i\sin\theta')$

Multiplying it is easily verified that
$zz'=rr'(\cos(\theta+\theta')+i\sin(\theta+\theta'))$

Thus we have $|zz'|=rr'=|z||z'|$

$\arg zz'=\arg z+\arg z'$  (mod$2\pi $)

\bigskip

\textbf{Exercise}

Prove by induction that $|z^n|=|z|^n$

$\arg z^n=n\arg z$   (mod$2\pi) \,\,\,\, n\epsilon$ {\bf N}

If $m=-n \,\,\,\, n\epsilon {\bf N}$

Then $z^mz^n=1$  So $|z^m||z^n|=1$

i.e. $|z^m||z|^n=1$

so $\ds |z^m|=\frac{1}{|z|^n}=|z|^m$

\bigskip

\textbf{Exercise}

Prove that if $m=-n \,\,\,\, n\epsilon$ {\bf N}

then $\arg z^m=m\arg z$   mod$2\pi$

The most important feature of the complex number system is that
not only does $x^2+1=0$ have a solution in {\bf C}, but all
polynomial equations have solutions in {\bf C}.

This fact was first given a complete proof by Gauss in 1799.

\bigskip

{\bf Fundamental Theorem of Algebra}

Let $\ds p(z)=a_0+a_1z+a_2z^2+...+a_nz^n \,\,\,\, a_i\epsilon${\bf
C}

Then the equation $p(z)=0$ has a solution in {\bf C}

It follows that if $c$ is a such solution then $\ds
p(z)=(z-c)(b_0+b_1z+...+b_{n-1}z^{n-1})$

\textbf{Exercise}

Try to prove this.

\textbf{Corollary}

$p(z)$ can be expressed as a product of $n$ linear factors.

$p(z)=a_n(z-c_1)(z-c_2)...(z-c_n)$ where some of the $c_i$ may be
equal.  Thus every polynomial has at most $n$ roots in {\bf C}.

Proof by induction is left as an exercise.

\bigskip

\textbf{Examples}

$\begin{array}{rcll} x^2+1 & & {\rm irreducible} & {\rm over\ }
{\bf R}\\ x^2+1 &=& (x+i)(x-i)& {\rm over\ } {\bf C}\\
x^3-x^2+2x-2 &=& (x-1)(x^2+2)& {\rm over\ } {\bf R}\\ x^3-x^2+2x-2
&=& (x-1)(x+i\sqrt2)(x-i\sqrt2)& {\rm over\ } {\bf C}\end{array}$

\bigskip

{\bf Complex conjugates}

Let $z=x+iy$.  Then we define the complex conjugate of $z$ to be
$\ol{z}$ or $z^*=x-iy$

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\textbf{Properties}

\begin{itemize}
\item[i)]
$\ol{z+w}=\ol{z}+\ol{w} \hspace{0.5in}$ direct verification

\item[ii)]
$\ol{zw}=\ol{z}\ol{w} \hspace{0.5in}$ direct verification

\item[iii)]
$\ol{z^n}=(\ol{z})^n \hspace{0.5in}$ from ii) by induction

\item[iv)]
If $p$ is a polynomial $p(z)=a_0+a_1z+...+a_nz^n$

$\ol{p(z)}=\ol{a}_0+\ol{a}_1\ol{z}+\ol{a}_2\ol{z}^2+...+\ol{a}_n\ol{z}^n$
\end{itemize}

This result is used to prove that if $z$ is a root of a polynomial
with real coefficients then $\ol{z}$ is also a root.  For in this
case if $a_i\epsilon$ {\bf R} then $\ol{a}_i=a_i$.

So $\ol{p(z)} =p(\ol{z})$

Thus if $p(z)=0 \,\,\,\, p(\ol{z})=0$ also.

So for a real polynomial all the complex roots have corresponding
conjugates.  Thus a real polynomial of odd degree must have at
least one real root.

\begin{itemize}
\item[v)]
$z\ol{z}=(x+iy)(x-iy)=x^2+y^2$

This is useful in such situations as

$\ds\frac{3+4i}{2-i}=\frac{(3+4i)(2+i)}{5}=\frac{2+11i}{5}$

\item[vi)]
$\ds\frac{z+\ol{z}}{2}=$ Re $z=x$

$\ds\frac{z-\ol{z}}{2}=i$ Im $z=iy$
\end{itemize}

\bigskip

{\bf De Moivre's Theorem}

$(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$

Proof by induction left as an exercise.

\bigskip

{\bf The roots of unity}

We consider the equation $z^n=1$.

If $z=r(\cos\theta+i\sin\theta)$

$z^n=r^n(\cos n\theta+i\sin n\theta)$

So $r^n=1$ which gives $r=1$, and $\cos n\theta=1$.

This gives $\ds \theta=0, \frac{2\pi}{n}, \frac{4\pi}{n}, ...,
\frac{(2n-2)\pi}{n}$

Thus the solutions of $z^n=1$ are

$\ds \cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n} \,\,\,\,
k=0,1,...n-1$ and these are all different.

For example take $n=3$ the roots of $z^3=1$ are

$\ds 1, \,\,\, \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}, \,\,\,
\cos{4\pi}{3}+i\sin\frac{4\pi}{3}$

$\ds 1, \,\,\, \frac{-1+\sqrt3i}{2}, \,\,\, \frac{-1-\sqrt3i}{2}$

They lie at the vertices of a regular triangle on the unit circle
in the complex plane.
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In general the $n$-th roots of unity lie at the vertices of a
regular $n$-gon.

\bigskip

\textbf{Exercise}

Let the cube roots of unity be denoted by $1,\, w_1,\, w_2$. Prove
that $w_1^2=w_2$ and $w_2^2=w_1.$

Prove that $w_1,\, w_1^2,\, w_1^3$ are all different, and $w_2,\,
w_2^2,\, w_2^3$ are all different, and each set is a permutation
of $1,\, w_1,\, w_2$. Investigate this situation for $n=5,8$ and
then see if you can make any general statements for the $n$-th
roots of unity.

Useful in this exercise might be:

\bigskip

{\bf Euler's formula}

Assuming the series for

$\ds e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

$\ds \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

$\ds \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

Replace $x$ by $ix$ in the first and separate the resulting series
into real and imaginary parts to verify that

$\ds e^{ix}=\cos x+i\sin x$

Then using $\ds e^{-ix}=\cos x-i\sin x$

We obtain $\ds \cos x=\frac{1}{2}(e^{ix}+e^{-ix})$

$\ds \hspace{0.8in} \sin x=\frac{1}{2i}(e^{ix}-e^{-ix})$

These look like the formula for hyperbolic functions.  Investigate
this connection further.

Euler's formula enables us to deal with the roots of unity more
concisely.

Since $\ds
\cos\frac{2k\pi}{n}+i\sin\frac{2k\pi}{n}=e^{i\frac{2k\pi}{n}}$

we can obtain them as follows

$z^n=1=e^{2\pi i}=e^{4\pi i}=...=e^{2k\pi i}$

so $z=e^\frac{2k\pi i}{n}$

The formula with DeMoivre's theorem is useful in summing series
and evaluating integrals.

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