\docname{sess6.5}
\doctitle{Polar Representation}

\mylhead
\myrhead

\title{\doctitle}
\maketitle

\section{The Complex Plane}

Complex numbers are represented geometrically by points in the plane:
the number $a+ib$ is represented by the point $(a,b)$ in Cartesian
coordinates. When the points of the plane are thought of as
representing complex numbers in this way, the plane is called the {\bf
  complex plane}.

By switching to polar coordinates, we can write any non-zero complex number in an alternative form. Letting as usual
\[x=r\cos\theta, \quad y=r\sin\theta\]
we get the {\bf polar form} for a non-zero complex number: assuming
$x+iy\neq 0$, 
\begin{equation}\label{eq6.5.8}
x+iy=r(\cos\theta+i\sin\theta).
\end{equation}

When the complex number is written in polar form, \[r=|x+iy| =
\sqrt{x^2 + y^2}. \qquad  \textit{ (absolute value, modulus)}\]
We call $\theta$ the \textit{polar angle} or the \textit{argument} of $x+iy$. In symbols, one sometimes sees:
\[\theta=\arrg(x+iy). \qquad \textit{ (polar angle, argument).}\]

The absolute value is uniquely determined by $x+iy$ but the polar
angle is not, since it can be increased by any integer multiple of
$2\pi$. (The complex number 0 has no polar angle.) To make $\theta$
unique, one can specify $$0\leq \theta<2\pi. \qquad \textit{(principal
value)}$$

This so-called principal value of the angle is sometimes indicated by writing $\Arrg(x+iy)$. For example,
\[\Arrg(-1)=\pi, \qquad \arrg(-1)=\pm\pi, \pm 3\pi, \pm 5\pi, \cdots\]

Changing between Cartesian and polar representation of a complex
number is essentially the same as changing between Cartesian and polar
coordinates: the same equations are used.

\examp{1} Give the polar form for: $-i, 1+i, 1-i, -1+i\sqrt{3}$.

\ans
\begin{align*}
-i&=i\sin\frac{3\pi}{2}&1+i&=\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})\\
-1+i\sqrt{3}&=2(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3})&1-i&=\sqrt{2}(\cos \frac{-\pi}{4}+i\sin \frac{-\pi}{4}).
\end{align*}

\section{Euler's Formula}

The abbreviation $\cis\theta$ is sometimes used for
$\cos\theta+i\sin\theta$; for students of science and engineering,
however, it is important to get used to the exponential form for this
expression:
\begin{equation}\label{eq6.5.9}
e^{i\theta}=\cos\theta+i\sin\theta\qquad \text{{\bf Euler's formula.}}
\end{equation}
Equation (\ref{eq6.5.9}) should be regarded as the \textit{definition} of the exponential of an imaginary power. A good justification for it is found in the infinite series:
\[e^t=1+\frac{t}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots\]
If we substitute $i\theta$ for $t$ in the series and collect the real and imaginary parts of the sum (remembering that $i^2=-1$, $i^3=-i$, $i^4=1$, $i^5=i$,
and so on), we get:
\begin{align*}
e^{i\theta}&=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right)+i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\right)\\
&=\cos\theta+i\sin\theta
\end{align*}
in view of the infinite series representations for $\cos\theta$ and $\sin\theta$.

Since we only know that the series expansion for $e^t$ is valid when
$t$ is a real number, the above argument is only suggestive --- it is
not a proof of (\ref{eq6.5.9}). What it shows is that Euler's formula
(\ref{eq6.5.9}) is formally compatible with the series expansions for
the exponential, sine, and cosine functions.

\section{Polar Representation}

Using the complex exponential, the polar representation (\ref{eq6.5.8}) is written:
\begin{equation}\label{eq6.5.10}
x+iy=re^{i\theta}.
\end{equation}

The most important reason for polar representation is that
multiplication of complex numbers is particularly simple when they are
written in polar form. Indeed, by using Euler's formula (\ref{eq6.5.9}) and the trigonometric addition formulas, it is not hard to show:
\begin{equation}\label{eq6.5.11}
e^{i\theta}e^{i\theta'}=e^{i(\theta+\theta')}.
\end{equation}
This gives another justification for the definition (\ref{eq6.5.9})
--- it makes the complex exponential follow the same exponential
addition rules as the real exponential. The law (\ref{eq6.5.11}) leads
to the simple rules for multiplying and dividing complex numbers
written in polar form:
\smallskip\\
{\bf multiplication rule}
\begin{equation}\label{eq6.5.12a}
r_1e^{i\theta} \cdot
r_2e^{i\theta'}=r_1r_2e^{i(\theta+\theta')}.
\end{equation}
\textit{To multiply two complex numbers, you multiply the absolute
  values and add the angles}\\
\smallskip\\
{\bf reciprocal rule}
\begin{equation}\label{eq6.5.12b}
\frac{1}{re^{i\theta}}=\frac{1}{r}e^{-i\theta};
\end{equation}
{\bf division rule}
\begin{equation}\label{eq6.5.12c}
\frac{re^{i\theta}}{r'e^{i\theta'}}=\frac{r}{r'}e^{i(\theta-\theta')}.
\end{equation}
\textit{To divide by a complex number, divide by its absolute value
  and subtract its angle.}

The reciprocal rule (\ref{eq6.5.12b}) follows from (\ref{eq6.5.12a}),
which shows that $$\frac{1}{r} e^{-i\theta} \cdot r e^{i \theta} = 1.$$

Using (\ref{eq6.5.12a}), we can raise $x + iy$ to a positive integer
power by first using $x + i y = re^{i\theta}$; the special case when
$r=1$ is called \textit{DeMoivre's formula}:\\
\begin{equation}\label{eq6.5.13}
(x+iy)^n = r^n e^{-n\theta};
\end{equation}
{\bf DeMoivre's formula}\\
\begin{equation}\label{eqdemoivre}
(\cos \theta + i\sin \theta)^n = \cos n\theta + i \sin n\theta.
\end{equation}

\examp{2} Express:
\begin{enumerate}[a)]
\item $(1+i)^6$ in Cartesian form;
\item $\displaystyle{\frac{1 + i\sqrt 3}{\sqrt 3 + i}}$ in polar form.
\end{enumerate}

\ans
\begin{enumerate}[a)]
\item Change to polar form, use (\ref{eq6.5.13}), then change back to Cartesian form:
\[(1+i)^6=(\sqrt{2} e^{i\pi/4})^6=(\sqrt{2})^6e^{i6\pi/4}=8e^{i3\pi/2}=-8i.\]

\item Changing to polar form, $\displaystyle{\frac{1 + i\sqrt 3}{\sqrt
      3 + i} = \frac{2e^{i \pi/3}}{2e^{i\pi/6}} = e^{i \pi/6}}$, using
  the division rule (\ref{eq6.5.12c}).
\end{enumerate}
You can check the answer to (a) by applying the binomial theorem to
$(1+i)^6$ and collecting the real and imaginary parts; to (b) by doing
the division in the Cartesian form then converting
the answer to polar form.

\hbcom{The next section is no longer in Mattuck's notes?  I believe there is a
  mathlet that illustrates this topic.}
\subsection{Combining pure oscillations of the same frequency.} The
equation which does this is widely used in physics and engineering; it
can be expressed using complex numbers:
\begin{equation}\label{eqosc}
A\cos\lambda t+B\sin\lambda t=C\cos(\lambda t+\phi),\hspace{1cm}\text{
  where } A+Bi=Ce^{i\phi};
\end{equation}
in other words, $C=\sqrt{A^2+B^2}$, $\phi=\tan^{-1} B/A$. To prove (\ref{eqosc}), we have:
\begin{align*}
A\cos\lambda t+B\sin\lambda t&=\Repart\left((A+Bi)\cdot(\cos\lambda t+i\sin\lambda t)\right)\\
&=\Repart(Ce^{i\phi}\cdot e^{i\lambda t})\\
&=\Repart(Ce^{\lambda t+\phi})=C\cos(\lambda t+\phi).
\end{align*}