\newlec

\hdc
Express the following in terms of polar coordinates:
\begin{enumerate}
\item $-3$
\item $1-i$
\end{enumerate}
\vskip3em
\soln
\begin{enumerate}
\item Note that $r = |-3| = 3$ and $\theta = \arctan(0/-3) = 0$. Then, since $-3$ lives between the second and third quadrant, you need to add $\pi$ to the previous answer. Thus $\theta = \pi$ and hence $-3 = 3\cis(\pi)$.
\ic{Make sure to note the addition of pi above.}
\item Note that $r = |1-i| = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$. Hence
\begin{align*}
1 - i &= \sqrt{2}\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \right)\\
&= \sqrt{2} (\cos(7\pi/4) + i\sin(7\pi/4))\\
&= \sqrt{2}\cis(7\pi/4)
\end{align*}
\end{enumerate}

\ic{This is the 10 minute mark.}

\newpage

\hdc
\begin{enumerate}
\item Write $\text{cis}(15\pi/6)$ in standard form.
\item Write $-3\sqrt{2} + 3\sqrt{6}i$ in polar form.
\end{enumerate}
\vskip3em
\soln 
\begin{enumerate}
\item $\cis(15\pi/6) = \cos(5\pi/2) + i\sin(5\pi/2) = i$.
\item Note that 
\begin{align*}
r &= |-3\sqrt{2} + 3\sqrt{6}i|\\
&= \sqrt{(-3\sqrt{2})^{2} + (3\sqrt{6})^{2}}\\
&=\sqrt{18 + 54}\\
&= \sqrt{72}\\
&=6\sqrt{2}
\end{align*}
Therefore, $-3\sqrt{2} + 3\sqrt{6}i = 6\sqrt{2}\left(\frac{-1}{2} + \frac{\sqrt{3}}{2}i \right) = 6\sqrt{2} \cis(2\pi/3)$ where the last equality holds since
\begin{center}
\includegraphics{PlaneAngleEx4.png}
\end{center}
\ic{This is the 20 minute mark}
\newpage
\thm (Polar Multiplication of Complex Numbers (PMCN)) If $z_{1} = r_{1} \cis(\theta_{1})$ and $z_{2} = r_{2} \cis(\theta_{2})$, then 
\[
z_{1}z_{2} = r_{1}r_{2}\cis(\theta_{1} + \theta_{2})
\]
\pf We have
\begin{align*}
z_{1}z_{2}&= r_{1}(\cos(\theta_{1}) + i\sin(\theta_{1})) r_{2}(\cos(\theta_{2}) + i\sin(\theta_{2}))\\
&= r_{1}r_{2}(\cos(\theta_{1})\cos(\theta_{2}) - \sin(\theta_{1})\sin(\theta_{2}) + i(\cos(\theta_{1})\sin(\theta_{2}) + \sin(\theta_{1})\cos(\theta_{2})))\\
&= r_{1}r_{2}(\cos(\theta_{1} + \theta_{2}) + i\sin(\theta_{1} +\theta_{2})) \\
&= r_{1}r_{2}\cis(\theta_{1} + \theta_{2})
\end{align*}
where in line 3 above, we used trig identities. This completes the proof. \qed
\end{enumerate}
\cor Multiplication by $i = \cos(\pi/2) + i\sin(\pi/2)$ gives a rotation by $\pi/2$.
\begin{center}
\includegraphics[scale=0.8]{PlaneAngleEx5.png}
\end{center}
\ex Using Polar Multiplication of Complex Numbers on $(\sqrt{6} + \sqrt{2}i)(-3\sqrt{2} + 3\sqrt{6}i)$ gives
\begin{align*}
(\sqrt{6} + \sqrt{2}i)(-3\sqrt{2} + 3\sqrt{6}i) &= 2\sqrt{2}\cis(\pi/6)\cdot6\sqrt{2}\cis(2\pi/3)\\
&= 24\cis(\pi/6 + 2\pi/3) &&\text{ By PMCN}\\
&= 24\cis(5\pi/6)\\
&= 24(-\sqrt{3}/2 + i/2)\\
&= -12\sqrt{3} + 12i
\end{align*}
\ic{This is the 30-35 minute mark.}
\thm (De Moivre's Theorem (DMT)) If $\theta \in \mathbb{R}$ and $n \in \mathbb{Z}$, then 
$$(\cos(\theta) + i\sin(\theta))^{n} = \cos(n\theta) + i\sin(n\theta)$$
or more compactly,
$$
\cis(\theta)^{n} = \cis(n\theta).
$$
\ic{Emphasize here that we want to use induction but need to reduce to the natural numbers first}
\pf First note that when $n=0$, we see that $(\cos(\theta) + i\sin(\theta))^{0} = 1$ and that $\cos(0\theta) + i\sin(0\theta) = 1$ so the statement holds.
For $n > 0$, we proceed by induction on $n$. For the base case, consider $n=1$. Then
\begin{align*}
(\cos(\theta) + i\sin(\theta))^{n} = \cos(\theta) + i\sin(\theta) = \cos(n\theta) + i\sin(n\theta).
\end{align*}
Now, assume that
\[
(\cos(\theta) + i\sin(\theta))^{k} = \cos(k\theta) + i\sin(k\theta)
\]
holds for some $k \in \mathbb{N}$. For the inductive step, note that
\begin{align*}
(\cos(\theta) + i\sin(\theta))^{k+1} &= (\cos(\theta) + i\sin(\theta))^{k}(\cos(\theta) + i\sin(\theta))\\
&= ( \cos(k\theta) + i\sin(k\theta))( \cos(\theta) + i\sin(\theta)) &&\text{Inductive hypothesis}\\
&= \cos((k+1)\theta) + i\sin((k+1)\theta) && \text{By PMCN}
\end{align*}
\vskip1em
For $n < 0$, we write $n = -m$ for some $m \in \mathbb{N}$. Then
\begin{align*}
\cis(\theta)^{n} &= \cis(\theta)^{-m}\\
&= (\cis(\theta)^{m})^{-1}\\
&= \cis(m\theta)^{-1}\\
&= \frac{\cos(m\theta) - i\sin(m\theta)}{\cos^{2}(m\theta) + \sin^{2}(m\theta)} && \text{Since $z^{-1} = \overline{z}/|z|^{2}$}\\
&= \cos(m\theta) - i\sin(m\theta)
\end{align*}
and $\cos(-m\theta) + i\sin(-m\theta) = \cos(m\theta) -i\sin(m\theta)$ since cosine is even and sine is odd. This completes the proof. \qed
\vskip1em
\cor If $z = r\cis(\theta)$ then $z^{n} = r^{n}\cis(n\theta)$.
\ic{This is the 50 minute mark.}

