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\thm Complex $n$th Roots Theorem (CNRT) Any nonzero complex number has exactly $n\in\mathbb{N}$ distinct $n$th roots. The roots lie on a circle of radius $|z|$ centred at the origin and spaced out evenly by angles of $2\pi/n$. Concretely, if $a = re^{i\theta}$, then solutions to $z^{n} = a$ are given by $z = \sqrt[n]{r}e^{i(\theta + 2\pi k)/n}$ for $k \in \{0,1,...,n-1\}$.
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\pf The proof is like the example yesterday and is left as additional reading. \qed
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\defn An $n$th root of unity is a complex number $z$ such that $z^{n} = 1$. These are sometimes denoted by $\zeta_{n}$.
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\ex $-1$ is a second root of unity (and a fourth root of unity and a sixth root of unity etc.)
\ic{This is the 10 minute mark; though likely the previous lecture spilled over to this lecture.}

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\hdc

Find all eighth roots of unity in standard form.

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\soln We want to solve $z^{8} = 1$. We know that $\{\pm1, \pm i\}$ are solutions. We can draw to find the rest:

\begin{center}
\includegraphics[scale=0.5]{CNRT2.png}
\end{center}

For another example, look at $z^{3} = 1$:
\begin{center}
\includegraphics[scale=0.5]{CNRT3.png}
\end{center}

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\ex Solve $z^{5} = -16\overline{z}$.
\ic{Get students to guess the total number of solutions. Also get them to find a solution by inspection. The answer is surprising!}
\soln This is a tricky problem. One could convert to polar coordinates but I prefer to reason as follows. If I can't solve the equation as written, maybe I can simplify by taking lengths on both sides.
\begin{align*}
|z^{5}| = |z|^{5} = |-16\overline{z}| = 16|\overline{z}|= 16|z|
\end{align*}
This gives $|z|^{5} = 16|z|$. Hence $|z|^{5} - 16|z| = 0$ giving $|z|(|z|^{4} - 16) = 0$. This gives either $|z| = 0$ which translates to $z=0$ or $|z|^{4} = 16$ which gives $|z| = 2$. So assuming that $z \neq 0$, we multiply the original equation by $z$ to yield
\[
z^{6} = -16z\overline{z} = -16|z|^{2} = -64
\]
but this question we solved before! Therefore,
\[
z \in \{0,\pm2i, \pm \sqrt{3} \pm i\}
\]
Thus, there are seven solutions!
\ic{This is the 40 minute mark; if you spilled over from the previous lecture, this is the 50 minute mark. Otherwise do the next problem (which is one we did before)}
\ex Solve $z^{6} + 2z^{3}  -3 = 0$. 
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\pf From before, we factored this to $(z^{3}-1)(z^{3}+3) = 0$ and thus $z^{3} = 1$ or $z^{3} = -3$. From CNRT, we see that the solutions to $z^{3} = 1 = \cos(0) + i\sin(0)$ are given by
\[
z \in \{e^{i \cdot 0}, e^{i \cdot 2\pi/3}, e^{i\cdot 4\pi /3}\}
\]
and solutions to $z^{3} = -3 = 3(\cos(\pi) + i\sin(\pi))$ are given by
\[
z \in \{\sqrt[3]{3}e^{i \cdot \pi/3}, \sqrt[3]{3}e^{i \cdot \pi}, \sqrt[3]{3}e^{i\cdot 5\pi /3}\}
\]
\begin{center}
\includegraphics[scale=0.5]{CNRT4.png}
\end{center}
This completes the question. \qed

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