\newlec

\hdc
Write $(\sqrt{3} - i)^{10}$ in standard form.
\vskip3em
\soln
Convert $\sqrt{3} - i$ to polar coordinates.
\begin{align*}
\sqrt{3} - i &= 2\left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)\\
&= 2\cis(-\pi/6)\\
&= 2\cis(11\pi/6)
\end{align*}
seen via the diagram
\begin{center}
\includegraphics[scale=0.8]{PlaneAngleEx6.png}
\end{center}
Lastly,
\begin{align*}
( 2\cis(11\pi/6))^{10} &= 2^{10}\cis(110\pi/6) &&\text{DMT}\\
&= 2^{10}\cis(55\pi/3)\\
&= 2^{10}\cis(9(2\pi) + \pi/3)\\
&= 2^{10}\cis(\pi/3)\\
&= 2^{10}\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\\
&= 2^{9} + 2^{9}\sqrt{3}i\\
&= 512 + 512\sqrt{3}i
\end{align*}
seen via the diagram
\begin{center}
\includegraphics[scale=0.8]{PlaneAngleEx7.png}
\end{center}
\ic{This is the 10-15 minute mark}
\newpage
\textbf{Complex Exponential Function}
\vskip1em
\defn For a real $\theta$, define
\[
e^{i\theta} := \cos(\theta) + i\sin(\theta) = \cis(\theta)
\]
\note Can write $z \in \mathbb{C}$ as $z = re^{i\theta}$ where $r = |z|$ and $\theta$ is an argument of $z$.
\vskip1em
\qn Why is this definition reasonable? While we can't prove the answer to this question, we can give convincing arguments.
\vskip1em
\noindent \textbf{Reason 1:} Exponential Laws Work! For $\theta,\alpha \in \mathbb{R}$ and $n \in \mathbb{N}$,
\begin{align*}
e^{i\theta} \cdot e^{i\alpha} &= e^{i(\theta + \alpha)} &&\text{PMCN}\\
(e^{i\theta})^{n} &= e^{in\theta} &&\text{DMT}
\end{align*}
\textbf{Reason 2:} Derivative with respect to $\theta$ makes sense.
\begin{align*}
\frac{d}{d\theta}(\cos(\theta) + i\sin(\theta)) &= -\sin(\theta) + i\cos(\theta)\\
&= i(\cos(\theta) + i\sin(\theta))\\
&= ie^{i\theta}
\end{align*}
\textbf{Reason 3:} Power series.
\begin{align*}
e^{x} &= 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\
\sin(x) &= x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - ...\\
\cos(x) &= 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - ...
\end{align*}
Using these and combining gives
\[
e^{ix} = \cos(x) + i\sin(x)
\]
Setting $\theta = \pi$ gives Euler's Formula:
\[
e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1
\]
\ic{This is the 25 minute mark.}
\ex Write $(2e^{11\pi/6})^{6}$ in standard form.
\vskip1em
\soln By exponent rules (DMT), we have
\begin{align*}
(2e^{11\pi/6})^{6} &= 2^{6}e^{11\pi i}\\
&= 2^{6}(\cos(11\pi) + i\sin(11\pi))\\
&= 2^{6}(-1 + 0i)\\
&= -64
\end{align*}
\ic{This is the 30 minute mark.}
\ex Solve $z^{6} + 2z^{3} - 3 = 0$
\vskip1em
\soln Factoring gives
\begin{align*}
0 = z^{6} + 2z^{3} - 3  = (z^{3}-1)(z^{3}+3)
\end{align*}
Hence $z^{3} = 1$ or $z^{3} = -3$.
\vskip1em
\qn Can we solve $z^{n} = w$ for a fixed $w \in \mathbb{C}$?
\vskip1em
\note We saw a case of this already with $n=2$ and $w = -r$. We'll delay the previous question until later.
\vskip1em
\ex Solve $z^{6} = -64$.
\vskip1em
\soln We already saw that $2e^{11\pi/6}$ was a solution. Note that $\pm 2i$ are two others found by inspection. How do we find all the solutions in general? The answer involves using polar coordinates. Write $z = re^{i\theta}$. Then
\begin{align*}
z^{6} = r^{6} e^{i6\theta} = -64
\end{align*}
Taking the modulus yields $|r|^{6}|e^{i6\theta}| = 64$. Since for any real $\alpha$, we have
\[
|e^{i\alpha}| = |\cos(\alpha) + i\sin(\alpha)| = \sqrt{\cos^{2}(\alpha) + \sin^{2}(\alpha)} = 1
\]
we see that $|r|^{6} = 64$ and hence $r = 2$ since $r$ is a positive real number.
\ic{This is the 40 minute mark.}
Hence, we see that $-64 = r^{6}e^{i6\theta} = 64e^{i6\theta}$ and so $e^{i6\theta} = -1$. Thus,
\begin{align*}
\cos(6\theta) + i\sin(6\theta) = -1 = \cos(\pi) + i\sin(\pi)
\end{align*}
Hence, this is true when $6\theta = \pi + 2\pi k$ for all $k \in \mathbb{Z}$. Solving for $\theta$ gives
\[
\theta = \frac{\pi + 2\pi k}{6} = \frac{\pi}{6} + \frac{\pi}{3}k
\]
Now, when do two values of $\theta$ coincide with the same complex point? Answer: When they differ by multiples of $2\pi$.
\vskip1em
\claim $\theta_{1} = \tfrac{\pi}{6} + \tfrac{\pi}{3}k_{1}$ and $\theta_{2} = \tfrac{\pi}{6} + \tfrac{\pi}{3}k_{2}$ are equal up to $2\pi$ rotations if and only if $k_{1} \equiv k_{2} \mod{6}$.
\vskip1em
\pf We have that 
\begin{align*}
\theta_{1} &= \theta_{2} + 2\pi m &&\text{for some $m \in\mathbb{Z}$}\\
\frac{\pi}{6} + \frac{\pi}{3}k_{1} = \frac{\pi}{6} + \frac{\pi}{3}k_{2} + 2\pi m\\
\frac{\pi}{3}k_{1} &= \frac{\pi}{3}k_{2} + 2\pi m\\
k_{1} &= k_{2} + 6m\\
k_{1} &\equiv k_{2} \mod{6}
\end{align*}
and each of the above steps are if and only if steps. This completes the proof of the claim.
\vskip1em
Hence $\theta = \tfrac{\pi}{6} + \tfrac{\pi}{3}k_{1}$ for $k_{1} \in \{0,1,2,3,4,5\}$. Thus,
\[
\theta \in \{\tfrac{\pi}{6},\tfrac{3\pi}{6},\tfrac{5\pi}{6},\tfrac{7\pi}{6},\tfrac{9\pi}{6},\tfrac{11\pi}{6}\}
\]
or rewritten as
\[
\theta \in \{\tfrac{\pi}{6} + \tfrac{\pi}{3}k_{1} : k_{1} \in \{0,1,2,3,4,5\}\}
\]
Therefore, $z = re^{i\theta} \in \{2e^{i(\pi/6 + \pi k/3)}: k \in \{0,1,2,3,4,5\}\}$.
\begin{center}
\includegraphics[scale=0.5]{CNRT1.png}
\end{center}
\ic{In all likelihood, the 50 minute mark is somewhere above. Carry through to lecture 39 as needed.}