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\homework{17: Complex Numbers}{Due Friday, Week 10}


Solve \textbf{three} of the following \textbf{six} problems.    As always, prove your claims/have fun! 


\begin{enumerate}
\item Let $z_1, z_2, z_3$ be three elements of $\mathbb{C}$ such that 
\begin{itemize}
\item $z_1 + z_2 + z_3 = 0$, and
\item $|z_1| = |z_2| = |z_3|.$
\end{itemize}
Show that these three points, as graphed in $\mathbb{C}$, form the vertices of an equilateral triangle.
\item A \textbf{$n^{\textrm{th}}$-root of unity} is a complex number $\omega \in \mathbb{C}$ such that $\omega^n = 1$.  For example, $-1$ and 1 are the only two \nth{2}-roots of unity, because they are the only two complex numbers $\omega$ such that $\omega^2 = 1$.
\begin{enumerate}
\item Find all of the fifth roots of unity.  (Hint: use polar co\"ordinates.  In other words, think of elements of $\mathbb{C}$ as numbers of the form $r e^{i\theta}$.)
\item Let $\omega$ be a $n^{\textrm{th}}$-root of unity not equal to 1.  Show that
\begin{align*}
\sum_{k=0}^{n-1} \omega^k = 0.
\end{align*}
(Hint: Draw all of these powers of $\omega$ on the complex plane for some small values of $n$, like 3 or 4. Graphically, why is this sum 0?)
\end{enumerate}
\item Take all of the $n$-th roots of unity $\omega_1, \ldots \omega_n$. Find a closed form for the product
\begin{align*}
\prod_{k=1}^n \omega_k.
\end{align*}
(It should be very simple.)

\item Consider the following game of ``Lights-Out:''
\begin{itemize}
\item We start with a circle of $n \geq 2$ lights $b_1,\ldots b_n$ , some of which are on and some of which are off.
\item The only available moves are the following: suppose that there is some divisor $d$ of $n$, and some bulb $b_i$ such that for any $k \equiv i \mod d$, $b_i$ has the same state as $b_k$.  Then we can switch the states of all of these lights at once.
\end{itemize}

Suppose your game starts with exactly one light on.  Can you turn all of the other bulbs on?



\item One of the best reasons for working in $\mathbb{C}$ is that every polynomial of degree $n$ has $n$ roots!  We don't have time to prove in this class, so we leave it here for the HW:
\begin{thm}
(The Fundamental Theorem of Algebra:)  Every complex polynomial $p(z)$ with degree $n$ has $n$ (possibly repeated) roots in the complex plane.  In other words, we can factor every degree $n$ polynomial into $n$ complex roots: i.e. we can always find constants such that
\begin{align*}
p(z) = C \cdot \prod_{k=1}^n (z - r_i).
\end{align*}
\end{thm}

\item As it turns out, there is a far stronger analogue to the above theorem, which says (basically) that we can factor not just polynomials, but entire power series into their roots!  This theorem is incredibly difficult to prove (you could easily spend a pair of quarters on complex analysis and not get to it,) so we state it without proof below:
\begin{thm}
Weierstrass Factorization Theorem: every complex power series $f(x) = \sum a_nz^n$ can be written in the form 
\begin{align*}
e^{g(z)} x^n\cdot \prod_{\text{all roots }r_k \text{ of f}} \left(1 - \frac{z}{r_k}\right),
\end{align*}
for some natural number $n$ and some other complex power series $g(x)$.  
\end{thm}
Basically, this says that we can separate any complex power series into its roots (the $(1 - \frac{z}{r_i})$ - parts and the $z^k$ part), times some $e^{g(z)}$-part that's never 0. 

If you apply this theorem to $\sin(z)$, you'll get the following formula:
\begin{align*}
\sin(z) = z \cdot \prod_{n\neq 0, n \in \mathbb{Z}}\left(1 - \frac{z}{\pi n} \right).
\end{align*}

Also do not try to prove this.  

Instead, do the following: \textbf{use} the $\sin(z)$ identity above to prove the following result:
\begin{align*}
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.
\end{align*}
\end{enumerate}


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