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\textbf{Math 425: Complex Variables}
\hfill
\textbf{Final Exam Solutions}

\bigskip

\begin{enumerate}

\item[1.] $1+i = 2^{1/2}e^{\pi i/4}$ so $(1+i)^{100} = 2^{50}e^{25\pi i}
= -2^{50}$.


\item[2.]  Since $\sum_{n=1}^\infty n z^{n-1} = \frac{d}{dz}\sum_{n=0}^\infty
z^n = \frac{d}{dz}\frac{1}{1-z} = \frac{1}{(1-z)^2}$, we find\newline
$\sum_{n=1}^\infty n\frac{i^n}{2^{n+1}} = \frac{i}{2^2}
\sum_{n=1}^\infty n\Big(\frac{i}{2}\Big)^{n-1} =
\frac{i}{2^2}\frac{1}{(1-(i/2))^2} = \frac{i}{(2-i)^2} = \frac{i}{3-4i} =
\frac{i(3+4i)}{3^2+4^2} = \frac{-4+3i}{25}$.


\item[3.] $u(x,y) = x^2 - y^2 + 2x + y$ is harmonic: $u_x = 2x + 2$, $v_y = -2y
+ 1$, so $u_{xx} + v_{yy} = 2 - 2 = 0$. to find $f(z) = u + i v$ we use the
Cauchy-Riemann equations.
Since $v_y = u_x = 2x +2$, we know $v = 2xy + 2y + g(x)$ where $g(x)$
is some function of $x$ (only). Then $v_x = 2y + g'(x) = -u_y = 2y - 1$,
and so $g'(x) = -1$ and $g(x) = -x$. Therefore, $f(z) = x^2 - y^2 + 2x + y
+ i(2xy + 2y -x) = (x+iy)^2 + 2(x + iy) - i(x + iy) = z^2 +(2-i)z$.


\item[4.] Since $f(z) = \frac{1}{z^2 + 4}$ is analytic in
$\C\setminus\{-2i,2i\}$, the power series for $f$ around $z =
3$ converges on the largest disk that fits in  the domain, $D(3;R)
\subset \C\setminus\{-2i,2i\}$. The radius of this disk, $R$, is thus
the distance from $3$ to~$2i$: $|3-2i| = \sqrt{3^2 + 2^2} = \sqrt{13}$.
The series cannot converge on any larger disk since $f$ has a pole
at~$2i$. Therefore, the radius of convergence is~$\sqrt{13}$.


\item[5.] By Cauchy's Theorem, the integral does not depend on
the path as long as the path stays in a simply connected
domain where the function is analytic, as it does in this
problem. In fact, if $F'(z) = f(z)$, then $\int_\gamma f(z)\,dz =
F(\gamma(b))-F(\gamma(a))$.  Therefore, $\int_\gamma \frac{z}{z^2+1}\,dz
= \frac{1}{2}\log(z^2+1)\Big|_{\gamma(0)}^{\gamma(\pi)} =
\frac{1}{2}\log(\pi^2+1)$.

\problemspace

\item[6.] By Cauchy's Formula, $\int_\gamma \frac{f(z)}{z-a}\,dz = 2\pi i
f(a)$, so $\int_\gamma \frac{e^z}{z+1}\,dz = 2\pi i e^{-1}$.

\problemspace

\item[7.] Integrating $1/(1+z) = 1 - z + z^2 - z^3 + \cdots$
gives $\log(1+z) = z - z^2/2 + z^3/3 - z^4/4 + \cdots$ and so
$\frac{\log(1+z)}{z^3} = \frac{1}{z^2} -  \frac{1}{2z} + \frac{1}{3} -
\frac{z}{4} + \cdots + \frac{(-z)^n}{n+3} + \cdots$

\problemspace

\item[8.] Since the integrand is a rational function with the degree
of the denominator at least 2 greater than the degree of the numerator,
the Residue Theorem implies that
  \begin{eqnarray*}
  \int_0^\infty \frac{x^2}{(x^2+1)^2}\,dx
  &=& \frac{1}{2}\int_{-\infty}^\infty \frac{x^2}{(x^2+1)^2}\,dx
  = \frac{1}{2}\, 2\pi i\, \Res\Big(\frac{z^2}{(z^2+1)^2};i\Big)
  = \pi i\, \frac{d}{dz} \Big[\frac{z^2}{(z+i)^2} \Big]_{z=i}\\
  &=&\pi i \Big[\frac{2z(z+i) - 2z^2}{(z+i)^3} \Big]_{z=i}
  =\pi i\, \frac{-4 + 2}{(2i)^3} = \frac{\pi}{4}
  \end{eqnarray*}

\problemspace

\item[9.] Let $p(z) = 3z^3 + 2z^2 + z + 7$. First try $n=1$ and let $q(z) =
-7$. Then for $|z| = 1$, $|p(z) + q(z)| = |3z^3 + 2z^2 + z| \le 3|z|^3 +
2|z|^2 + |z| = 6 < 7  = |q(z)|$. By Rouch\'e's Theorem, the number of
zeros of $p$ in $|z|\le 1$ equals the number of zeros of $q$ in $|z|\le
1$, that is, \emph{none}. Now try $n=2$ and let $q(z) = -3z^3$. Then
for $|z| = 2$, $|p(z) + q(z)| = |2z^2 + z + 7| \le 2|z|^2 + |z| + 7 =
17 < 24 = |q(z)|$.  By Rouch\'e's Theorem, the number of zeros of $p$
in $|z|\le 2$ equals the number of zeros of $q$ in $|z|\le 2$, that is,
\emph{three}. These zeros account for all the zeros of $p$ since $p$
has degree three. Therefore $n=2$ is the \emph{smallest} positive integer
such that all the zeros of $p$ lie in $|z|\le n$.

\problemspace

\item[10.] Let $T$ be the FLT such that $T(1) = 0$, $T(1+i) = 1$, and
$T(\infty) = \infty$. Then $T(z) =\frac{z-1}{(1+i)-1} = i(1-z)$, and $T$
must take the line $\Re(z) = 1$ into the real axis. Now let $S$ be the
FLT such that $S(-1) = 0$, $S(i) = 1$, and $S(1) = \infty$. Then
$S(w) = \frac{w+1}{w-1}\frac{i-1}{i+1} = i\frac{w+1}{w-1}$ and $S$ must
take the unit circle into the real axis. It follows that $F = S^{-1}\circ
T$ is a FLT that takes the line $\Re(z) = 1$ into the unit circle. To find
$w = F(z) = S^{-1}(T(z))$ we solve $S(w) = T(z)$ for $w$:
$i\frac{w+1}{w-1} = i(1-z) \Leftrightarrow w = \frac{z-2}{z} = F(z)$.
To see that $F$ maps $\Re(z) > 1$ into the disk, note that
$|F(z)| = \frac{(x-2)^2 + y^2}{x^2 + y^2} < 1 \Leftrightarrow
x^2 - 4x + 4 + y^2 < x^2 + y^2 \Leftrightarrow x > 1$.

\end{enumerate}

\end{document}
