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\prof{Peter Cholak and Juan Migliore}
\class{Math 222}
\quizdate{Friday, January 26}
\pagestyle{empty}

\begin{document}


\heading

\noindent {\bf Solutions}

\bigskip

Be sure to carefully write up your answers.  It is suggested that you
first write out a draft of your proposed questions and then carefully
rewrite that draft to get your final version.  You do \emph{not} have
to write the answers on this sheet of paper, but please staple all
sheets together before turning your quiz in.\\

\begin{problem}
  Show by induction that $n^4 -4n^2$ is divisible by $3$ for all $n
  \geq 1$.
\end{problem}

First check it for $n=1$: $1 - 4 = -3$ is divsible by 3.  Now assume that
$k^4 - 4k^2$ is divisible by 3 and consider 
\[
\begin{array}{rcl}
(k+1)^4 - 4(k+1)^2 & = & [k^4 + 4k^3 + 6 k^2 + 4k + 1] - 4[k^2 + 2k+1] \\
& = & [k^4 - 4k^2 ] + [4k^3 - 4k] + 6k^2 - 3 \\
& = & [k^4 - 4k^2 ] + [4(k-1)(k)(k+1)] + [6k^2 - 3].
\end{array}
\]
The first bracketed expression is divisible by 3 because of the inductive
hypothesis.  The second bracketed expression is divisible by 3 because it
is 4 times the product of three consecutive integers.  The last bracketed
expression is obviously divisible by 3.  Hence the whole expression is
divisible by 3.
\bigskip


\begin{problem}
In the ``real'' world, we are used to the fact that a power 
of a  non-zero number is again non-zero.  We will later see that this
fact is not true in all worlds.  Show that the complex numbers do
behave normally in this regard.  Precisely, let $z$ be an
arbitrary  complex number and suppose that we have $z^k = 0$ for some
positive integer $k$.  Show that $z$  must be zero.  (You may use the
fact that you know it to be true for real numbers.)
\end{problem}

If $z^k = 0$ then $0 = |z^k| = |z|^k$.  But $|z|$ is a real number, and
we know that a power of a non-zero real number is non-zero.  Hence we
have $|z| = 0$.  But we've shown that this implies $z=0$.

\bigskip

\begin{problem}
  Find {\em all} the cube roots of $-8i$.  Write them in the the form
  $a+bi$.  Be sure to show your work.
\end{problem}

Let $z = -8i$.  Then $\hbox{arg}(z) = \frac{3\pi}{2}$ and $|z| = 8$. We 
know that $\sqrt[3]{z}$ is a set with three elements.  Each of the
elements has modulus equal to $8^{\frac{1}{3}} = 2$.  Let $w$ be the
first element in $\sqrt[3]{z}$.  Then $\hbox{arg}(w) =
(\frac{3\pi}{2})/3 = \frac{\pi}{2}$.  The other two roots then have
arguments $\frac{\pi}{2} + \frac{2\pi}{3}$ and $\frac{\pi}{2} +
\frac{4\pi}{3}$, i.e.\ $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$.  So
\[
\begin{array}{rcl}
\sqrt[3]{-8i} & = & \left \{
2(\cos(\frac{\pi}{2}) + i \sin(\frac{\pi}{2})) , 
2(\cos(\frac{7\pi}{6}) + i \sin(\frac{7\pi}{6})) ,
2(\cos(\frac{11\pi}{6}) + i \sin(\frac{11\pi}{6})) 
\right \} \\
& = & \left \{ 2i, -\sqrt{3}-i, \sqrt{3} -i
\right \}.
\end{array}
\]

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