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\begin{document}
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\textbf{MATH 425: Complex Variables}
\hfill
\textbf{Final Exam Solutions, Spring 2003}

\bigskip

\begin{enumerate}

\setlength{\itemsep}{10pt}

\item[1.]
$\big(z + \frac{1}{z}\big)^2 = 1 \Rightarrow (z^2 + 1)^2 = z^2 \Rightarrow
z^4+z^2+1=0 \Rightarrow z^2 = \frac{-1\pm i\sqrt{3}}{2}= e^{\pm 2\pi i/3}
\Rightarrow z = \pm e^{\pm \pi i/3} = \pm \frac{1\pm i\sqrt{3}}{2}$

\item[2.] $e^z = e^x\cos(y) + ie^x\sin(y)$, $z = x + iy$.
This function is analytic because $u = e^x\cos(y)$ and $v = e^x\sin(y)$
satisfy the Cauchy-Riemann equations: $u_x = e^x\cos(y) = v_y$ and $u_y =
-e^x\sin(y) = -v_x$.

\item[3.] $u = x^3 - (3y^2+1)x$ is harmonic: $u_x = 3x^2 - 3y^2 - 1$,
$u_{xx} = 6x$, $u_y = -6xy$, $u_{yy} = -6x$, so $u_{xx} + u_{yy} = 6x - 6x
= 0$. The harmonic conjugate $v$ must satisfy the Cauchy-Riemann equations:
$v_y = u_x = 3x^2-3y^2 - y$ so $v = 3x^2y-y^3 - y + g(x)$ for some function
$g(x)$, and  $v_x = 6xy + g'(x) = -u_y = 6xy$ implies that $g'(x) = 0$ so
$g(x)$ is a constant, $c$. Thus $v = 3x^2y - y^3 - y + c$.

\item[4.] The line segment is paramterized by $z = 1-2t + i t$, $0 \le t
\le 1$, so\newline $\int_\CC \overline{z}\,dz = \int_0^1(1-2t-it)(-2+i)dt = (t-t^2
- it^2/2)(-2+i)\big|_0^1 = (-i/2)(-2+i) = 1/2 + i$.

\item[5.] Use the branch of the logarithm defined by $\log(z) = \ln|z| +
i\theta$ where $-3\pi/2 < \theta < \pi/2$, so an anti-derivative of
$z^{1/3} = e^{\log(z)/3}$ is defined on a domain that contains the
contour~$\CC$. Thus, $\int_\CC\sqrt[3]{z}\,dz
= \frac{3}{4}z^{4/3}\big|_{-1}^1
= \frac{3}{4}\big(1 - (-1)^{4/3}\big) 
= \frac{3}{4}\big(1 - e^{-4\pi i/3}\big) 
= \frac{3}{4}\big(1 - \frac{-1+i\sqrt{3}}{2}\big)
= \frac{9}{8} - i\frac{3\sqrt{3}}{8}$


\item[6.]
$\sum_{n=1}^\infty n^2z^n
= z \sum_{n=1}^\infty n^2z^{n-1}
= z \frac{d}{dz}\sum_{n=1}^\infty nz^n
= z \frac{d}{dz}\big[z\sum_{n=1}^\infty nz^{n-1}\big]
= z \frac{d}{dz}\big[z\frac{d}{dz}\sum_{n=0}^\infty z^n\big]
= z \frac{d}{dz}\big[z\frac{d}{dz}\frac{1}{1-z}\big]
= z \frac{d}{dz}\big[\frac{z}{(1-z)^2}\big]
= \frac{z(1+z)}{(1-z)^3}$

\item[7.]
$f(z) = \frac{1}{z^2-3z+2}
= \frac{1}{(z-1)(z-2)}
= -\frac{1}{z-1} + \frac{1}{z-2}
= -\frac{1}{z}\frac{1}{1-1/z} -\frac{1}{2} \frac{1}{1-z/2}$. Since $|1/z|<1$
and $|z/2|<1$, these functions can be expanded using geometric series to
give the Laurent series for $f(z)$:
$f(z)
= -\frac{1}{z}\sum_{n=0}^\infty\frac{1}{z^n} -\frac{1}{2}
\sum_{n=0}^\infty\frac{z^n}{2^n}
= -\sum_{n=1}^\infty\frac{1}{z^n} -
\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}$

\item[8.]
$\int_\CC \frac{e^z}{z^3+z^4}\,dz = 2\pi i\big[ \Res_0 + \Res_{-1}\big]
= 2\pi i \big[ \frac{1}{2}\frac{d^2}{dz^2}\frac{e^z}{z+1}\big|_{z=0} +
\frac{e^z}{z^3}\big|_{z=-1}\big] = 2\pi i [\frac{1}{2} - e^{-1}] = \pi
i(1-2/e)$.

\item[9.] Let $\CC$ be the usual contour along the $x$-axis from $-R$ to
$R$ and around the upper half of the circle $\CC_R$ of radius~$R$. Since
$f(z) = \frac{e^{iz}}{z^2+1}$ has simple poles at $z=\pm i$,
$\int_\CC f(z)\,dz = 2\pi i \Res_i = 2\pi i \frac{e^{iz}}{z+i}\big|_{z=i} =
\pi e^{-1}$. Now $\int_\CC f(z)\,dz = \int_{-R}^R f(x)\,dx + \int_{\CC_R}
f(z)\,dz = \int_{-\infty}^\infty \frac{e^{ix}}{x^2+1}\,dx + 0$ since
$\lim_{R\to \infty} \int_{\CC_R} f(z)\,dz = 0$. Therefore,
$\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1} \,dx +i \int_{-\infty}^\infty
\frac{\sin(x)}{x^2+1} \,dx = \pi/e$ and hence
$\int_0^\infty\frac{\cos(x)}{x^2+1} \,dx = \pi/(2e)$.

\item[10.]
Let $p(z) = z^5 - 3z^3 + 1$. To count the number of zeros of $p(z)$ in the
disk $|z| \le 1$, let $f(z) = -3z^3$ and $g(z) = z^5 + 1$. Then, for $|z| =
1$, we find $|g(z)| \le |z|^5 + 1 = 2 < |f(z)| = 3$. By Rouch\'e's
Theorem, the number of zeros of $p(z) = f(z)+g(z)$ in $|z|\le 1$ is the
same as the number of zeros of $f(z)$ in $|z|\le 1$, i.e.,~3.
To count the number of zeros of $p(z)$ in the
disk $|z| \le 2$, let $f(z) = z^5$ and $g(z) = -3z^3+1$. Then, for $|z| =
2$, $|g(z)| \le 3|z|^3 + 1 = 25 < |f(z)| = 32$. Therefore, the number of
zeros of $p(z) = f(z) + g(z)$ in $|z| \le 2$ is the same as the number of
zeros of $f(z)$ in $|z|\le 2$, i.e.,~5. Thus, the number of zeros of $p(z)$
in $1 \le |z| \le 2$ is $5-3 = 2$.

\end{enumerate}

\end{document}
