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

\bigskip

\begin{enumerate}

\item (a)
Let $u_n = n!|z|^{2n}/\sqrt{(2n)!}$. Then
$\lim_{n\to\infty}u_{n+1}/u_n = \lim_{n\to\infty}
(n+1)|z|^2/\sqrt{(2n+2)(2n+1)} = |z|^2/2$. By the Ratio Test, the
series converges absolutely for $|z|^2/2 < 1$, i.e., $|z| < \sqrt{2}$,
and diverges for $|z| >\sqrt{2}$. Therefore, the radius of convergence
is $\sqrt{2}$.

(b) $\sum_{n=0}^\infty (z-1)^n = \frac{1}{1-(z-1)} = \frac{1}{2-z}$,
so $\sum_{n=1}^\infty n(z-1)^n = (z-1)\sum_{n=1}^\infty n(z-1)^{n-1} =
(z-1)\frac{d}{dz} \sum_{n=0}^\infty (z-1)^n =
(z-1)\frac{d}{dz} \Big\{\frac{1}{2-z}\Big\} = \frac{z-1}{(z-2)^2}$. Then
$\sum_{n=1}^\infty n^2(z-1)^n = (z-1)\sum_{n=1}^\infty n^2(z-1)^{n-1} =
(z-1)\frac{d}{dz} \sum_{n=1}^\infty n(z-1)^n =
(z-1)\frac{d}{dz}\Big\{\frac{z-1}{(z-2)^2}\Big\}= -\frac{z(z-1)}{(z-2)^3}$.

(c) $\frac{z^m}{1-z^n} = z^m\sum_{k=0}^\infty \big(z^n\big)^k
= \sum_{k=0}^\infty z^{m + kn} = z^m + z^{m+n} + z^{m+2n} + \cdots$. The
radius of convergence is~1.

\problemskip

\item (a) $\frac{e^{iz}}{z^2+1} = \frac{f(z)}{z-i}$ where $f(z) =
\frac{e^{iz}}{z+i}$. By Cauchy's Formula (or direct calculation of the line
integral) $\int_{|z-i|=1}\frac{e^{iz}}{z^2+1}\,dz =\int_{|z-i|=1}
\frac{f(z)}{z-i}\,dz = 2\pi i f(i) = 2\pi i \frac{e^{-1}}{2i} =
\frac{\pi}{e}$.

(b) $\frac{1}{a + b\cos(\theta)} = \frac{1}{a + b(z + z^{-1})/2}
= \frac{2z}{bz^2 + 2az + b}$ where $z = e^{i\theta}$. Since $dz =
ie^{i\theta}d\theta$, we find that $d\theta = \frac{dz}{iz}$ and
$\int_0^{2\pi}\frac{1}{a + b\cos(\theta)} = \int_{|z|=1} \frac{2z}{bz^2
+ 2az + b}\frac{dz}{iz} = \frac{1}{i}\int_{|z|=1}\frac{2}{bz^2 + 2az +
b}\,dz$.  The denominator factors, $bz^2 + 2az + b = b(z-r_+)(z-r_-)$
where $r_\pm = -\frac{a}{b} \pm \sqrt{\frac{a^2}{b^2} - 1}$. Note that
 $\Big(\frac{a}{b} -1\Big)^2 = \frac{a^2}{b^2} - 2\frac{a}{b} + 1  <
\frac{a^2}{b^2} - 1$ because $\frac{a}{b} > 1$. Therefore, $-1 <
r_+ = -\frac{a}{b}+\sqrt{\frac{a^2}{b^2} - 1} < 0$, and so $r_+$
lies inside the circle $|z|=1$. Obviously, $r_- < -1$, so $r_-$
lies outside the cirlce $|z| = 1$. Writing $\frac{2}{bz^2 + 2az +
b} = \frac{f(z)}{z-z_+}$ where $f(z) = \frac{2}{b(z-z_-)}$, we may
apply Cauchy's Formula to obtain $\int_0^{2\pi}\frac{d\theta}{a +
b\cos(\theta)} = \frac{1}{i}\int_{|z|=1}\frac{f(z)}{z-z_+}\,dz
= 2\pi i\frac{1}{i} f(z_+) = 2\pi \frac{2}{b(z_+-z_-)} = 2\pi
\frac{2}{b\Big(2\sqrt{\frac{a^2}{b^2}-1}\Big)} = \frac{2\pi}{\sqrt{a^2
- b^2}}$.

(c) By defining $\log(z) = \log|z| + i\arg(z)$ with $-\pi/2 <
\arg(z) < 3\pi/2$, the function $f(z) = \log(z)/z$ is analytic in
$D = \C\setminus\{iy\,|\, y\le 0 \}$, and has an antiderivative
$F(z) = \log(z)^2/2$. Since $\gamma$ lies in the simply-connected
domain~$D$, Cauchy's Theorem implies that $\int_\gamma
\frac{\log(z)}{z}\,dz = F(\gamma(\pi)) - F(\gamma(0)) = (\log(-1)^2 -
\log(1)^2)/2 = (\pi i)^2/2 = -\pi^2/2$.

\problemskip

\item (a) Integrating $1/(1-z) = 1 + z + z^2 + \cdots$ we get
$\log(1-z) = -z - z^2/2 - z^3/3 - \cdots$ and $\log(1+z) = z - z^2/2 + z^3/3
- \cdots$ for $|z|<1$. Then
  \begin{eqnarray*}
  (1+z)^z &=& e^{z\log(1+z)} = e^{z^2 - z^3/2 + z^4/3 - \cdots}\\
  &=& 1+\big(z^2-\frac{1}{2}z^3+\frac{1}{3}z^4-\cdots\big)+
    \frac{1}{2}\big(z^2-\frac{1}{2}z^3+\frac{1}{3}z^4-\cdots\big)^2+\cdots\\
  &=& 1+z^2-\frac{1}{2}z^3+\frac{5}{6}z^4+\cdots
  \end{eqnarray*}
and
  \begin{eqnarray*}
  (1-z)^z &=& e^{z\log(1-z)} = e^{-z^2 - z^3/2 - z^4/3 - \cdots}\\
  &=& 1+\big(-z^2-\frac{1}{2}z^3-\frac{1}{3}z^4-\cdots\big)+
    \frac{1}{2}\big(-z^2-\frac{1}{2}z^3-\frac{1}{3}z^4-\cdots\big)^2+\cdots\\
  &=& 1-z^2-\frac{1}{2}z^3+\frac{1}{6}z^4+\cdots
  \end{eqnarray*}
So, $f(z) = (1+z)^z - (1-z)^z = 2z^2 + \frac{2}{3}z^4 + \cdots =
z^2\big(2 + \frac{2}{3}z^2 + \cdots\big)$ has a zero of order~2 at $z=0$.
\newline
(Note: This solution is much more than is needed. It would be enough
to show $f(0) = f'(0) = 0$ and $f''(0)\ne0$ by directly calculating
derivatives.)

(b) By Theorem~1, p.123, $f^{(n)}(z_0) = \frac{n!}{2\pi
i}\int_{|z-z_0| = r} \frac{f(z)}{(z - z_0)^{n+1}}\,dz$. Substituting
$z = z_0+ re^{i\theta}$, $dz = ire^{i\theta}d\theta$,
$\big|f^{(n)}(z_0)\big| \le \frac{n!}{2\pi}\int_0^{2\pi}
\frac{|f(z_0+re^{i\theta})|}{r^{n+1}|e^{(n+1)i\theta}|}r|ie^{i\theta}|\,d\theta
\le \frac{n!}{2\pi}\int_0^{2\pi} \frac{M}{r^n}\,d\theta
= \frac{n!}{2\pi}\frac{M}{r^n}2\pi = \frac{n!M}{r^n}$.

(c) Let $f(z)$ be a bounded entire function, say $f(z) \le M$ for all~$z$.
Let $z_0$ be an arbitrary point in the plane. By part (b), 
$f'(z_0) \le \frac{M}{r}$ for any~$r$. Therefore, $f'(z_0) = 0$. Since
$z_0$ was arbitrary, $f'(z) = 0$ for all $z$, and hence $f$ is constant.

\problemskip

\item (a) $f(z) = \frac{e^{2z}-1}{e^{3z}-e^{-z}}$ has singularities where
the denominator is zero: $e^{3z} - e^{-z} = 0$. This occurs when
$e^{3z} = e^{-z}$, i.e., $e^{4z} = 1$. Therefore, the singularities of $f$
are at the points $2n\pi i/4 = n\pi i/2$, $n\in\Z$. Note that
$f(z) =\frac{e^{2z}-1}{e^{-z}(e^{4z}-1)} =
\frac{e^z(e^{2z}-1)}{(e^{2z}+1)(e^{2z}-1)} = \frac{e^z}{e^{2z}+1}$,
which shows that the singularities at $(2k)\pi i/2 = k\pi i$ are
removable. In fact, $f(k\pi i) = e^{k\pi i}/(e^{2k\pi i}+1) = (-1)^k/2$.
To see that $f$ has poles of order~1 at the points $z_k = (2k+1)\pi
i/2$, we need only check for the order of the zero of the denominator
(the numerator, $e^z$, is never zero): Let $g(z) = e^{2z}+1$. Then
$g(z_k) = e^{(2k+1)\pi i}+1 = 0$, and $g'(z_k) = 2e^{(2k+1)\pi i/2} =
2(-1)^ki \ne 0$, so $g$ has a zero of order~1 at $z_k$.

(b) If $0<|z|<1$, then $f(z) = -\frac{1}{z}\cdot\frac{1}{1-z^2} =
-\frac{1}{z}\big(1+z^2+z^4+z^6\cdots\big) = -\frac{1}{z}-z-z^3-z^5-\cdots$.
\newline
If $|z|>1$, then $\Big|\frac{1}{z}\Big|<1$, so $f(z) =
\frac{1}{z^3}\cdot\frac{1}{1-1/z^2} =\frac{1}{z^3}\big(1 + \frac{1}{z^2}
+ \frac{1}{z^4} + \cdots\big) = \frac{1}{z^3} + \frac{1}{z^5} +
\frac{1}{z^7} + \cdots$.

(c) $f(z) = \frac{z^m}{z^n+1}$ has a simple pole at $z_0 = e^{\pi i/n}$
because the numerator is not zero at $z_0$,
and the denominator $g(z) = z^n+1$ satisfies $g(z_0) = e^{\pi i}+1 = 0$
while $g'(z_0) = ne^{(n-1)\pi i/n} = -ne^{-\pi i/n} \ne 0$. Therefore,
$\Res(f;z_0) = \lim_{z\to z_0} \frac{(z-z_0)z^m}{z^n+1} =
\frac{z_0^m}{nz_0^{n-1}} = \frac{1}{n}e^{(m-n+1)\pi i/n} =
-\frac{1}{n}e^{(m+1)\pi i/n} = -\frac{1}{n}\cos((m+1)\pi /n)
-\frac{i}{n}\sin((m+1)\pi/n)$.

\problemskip

\item (a) Let $f(z) = \frac{z^2}{(z^2+9)(z^2+4)^2} = 
\frac{z^2}{(z-3i)(z+3i)(z-2i)^2(z+2i)^2}$. Then $f$ has
simple poles at $z = \pm 3i$ and poles of order~2 at $z = \pm
2i$. Since the degree of the denominator is 2 more
than the degree of the numerator, the proposition on p.155
gives $\int_{-\infty}^\infty \frac{x^2}{(x^2+9)(x^2+4)^2}\,dx
= 2\pi i\big(\Res(f;3i) + \Res(f;2i)\big)$. Now,
  \[ \Res(f;3i) = \frac{z^2}{(z+3i)(z^2+4)^2}\Big|_{z=3i} =
  \frac{-9}{(6i)(-5)^2} = \frac{3}{50}i = \frac{12}{200}i \]
and
  \begin{eqnarray*}
  \Res(f;2i)
  &=& \Big[\frac{z^2}{(z^2+9)(z+2i)^2}\Big]'\Big|_{z=2i} =
    \Big[\frac{2z}{(z^2+9)(z+2i)^2} - \frac{2z^3}{(z^2+9)^2(z+2i)^2} -
    \frac{2z^2}{(z^2+9)(z+2i)^3}\Big]\Big|_{z=2i}\\
  &=& \frac{4i}{5(4i)^2}
    - \frac{2(2i)^3}{(5)^2(4i)^2} - \frac{2(2i)^2}{(5)(4i)^3} =
    -\frac{i}{20}-\frac{i}{25} + \frac{i}{40} = -\frac{13}{200}i
  \end{eqnarray*}
Therefore, $\int_{-\infty}^\infty \frac{x^2}{(x^2+9)(x^2+4)^2}\,dx =
2\pi i\Big(\frac{12}{200}i -\frac{13}{200}i\Big) = \frac{\pi}{100}$.

(b) The analytic function $f(z) = \frac{e^{iz}}{z^2+4z+5} =
\frac{e^{iz}}{(z+2-i))(z+2+i)}$ has simple poles at $z = -2+i$ and $z
= -2-i$. Let $\gamma = \gamma_1 + \gamma_2$ be the usual semicircular
curve defined by $\gamma_1(x) = x$, $-R \le x \le R$, and $\gamma_2(t)
= Re^{it}$, $0 \le t \le \pi$. Let $R$ be large enough so that $-2+i$
is inside $\gamma$. Then the Residue Theorem gives
   \[ \int_\gamma f(z)\,dz = 2\pi i \Res(f; -2+i) = 2\pi i
  \Big[\frac{e^{iz}}{z+2+i}\Big]\Big|_{z=-2+i} = 2\pi i
  \frac{e^{-1-2i}}{2i} = \frac{\pi}{e}e^{-2i} = \frac{\pi}{e}(\cos(2)
  - i\sin(2)) \]
On the other hand, $\int_\gamma f(z)\,dz = \int_{\gamma_1}
f(z)\,dz + \int_{\gamma_2} f(z)\,dz = \int_{-R}^R f(x)\,dx +
\int_0^{\pi}f(Re^{it})iRe^{it}\,dt$.  Now,
  \[ \Big|\int_0^{\pi}f(Re^{it})iRe^{it}\,dt\Big|
  \le \int_0^{\pi}|f(Re^{it})| R\,dt \le
  \int_0^{\pi}\frac{e^{-R\sin(t)}R}{(R-\sqrt{5})^2}\,dt \le
  \frac{R}{(R-\sqrt{5})^2}\pi \longrightarrow0,\ \mbox{as}\ R\to\infty \]
Letting $R\to\infty$, we conclude that
$\int_{-\infty}^{\infty}\frac{e^{ix}}{x^2+4x+5}\,dx
= \int_{-\infty}^{\infty} f(x)\,dx = \int_{\gamma}
f(z)\,dz = \frac{\pi}{e}(\cos(2)-i\sin(2))$. Therefore,
$\int_{-\infty}^{\infty}\frac{\sin(x)}{x^2+4x+5}\,dx
= -\frac{\pi}{e}\sin(2)$ and
$\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2+4x+5}\,dx =
\frac{\pi}{e}\cos(2)$.

(c) Let $f(z) = \frac{\log(z)}{z^4+1}$ where $\log(z) = \log|z| + i\arg(z)$
with $0 < \arg(z) < 2\pi$. Then $f$ has simple poles at the points
$z_n = e^{(2n+1)\pi i/4}$, $n=0,1,2,3$.
The residues of $f$ at these poles may be computed by $\lim_{z\to z_n}
\frac{(z-z_n)\log(z)}{z^4+1} = \frac{\log(z_n)}{4z_n^3}$.
Let $\gamma = \gamma_1 + \gamma_2 + \gamma_3 - \gamma_4$, where
$\gamma_1(x) = x$, $\epsilon \le x \le R$,
$\gamma_2(t) = Re^{it}$, $0 \le t \le \pi$,
$\gamma_3(x) = x$, $-R \le x \le -\epsilon$, and
$\gamma_4(t) = \epsilon e^{it}$, $0 \le t \le \pi$.
  \[ \resizebox{!}{1in}{\includegraphics{path.eps}} \]
Choose $R$ large enough and $\epsilon$ small enough so that $z_0$ and $z_1$
are inside $\gamma$.  By the Residue Theorem, 
  \begin{eqnarray*}
  \int_\gamma f(z)\,dz &=& 2\pi i(\Res(f;z_0) + \Res(f;z_1))
    = 2\pi i\Big(\frac{\log(e^{\pi i/4})}{4e^{3\pi i/4}} 
    +\frac{\log(e^{3\pi i/4})}{4e^{9\pi i/4}}\Big)\\
  &=& 2\pi i\Big(\frac{\pi i/4}{4e^{3\pi i/4}} 
    +\frac{3\pi i/4}{4e^{9\pi i/4}}\Big)
    = -\frac{\pi^2}{8} \Big(e^{-3\pi i/4} 
    +3e^{-\pi i/4}\Big)\\
  &=& -\frac{\pi^2}{8} \Big(\frac{-1-i}{\sqrt{2}} + 3\frac{1-i}{\sqrt{2}}\Big)
    = -\frac{\pi^2}{8} \Big(\frac{2-4i}{\sqrt{2}}\Big)\qquad(*)
  \end{eqnarray*}
On the other hand,
  \begin{eqnarray*} \int_\gamma f(z)\,dz
  &=& \int_{\gamma_1}f(z)\,dz +\int_{\gamma_2}f(z)\,dz
    +\int_{\gamma_3}f(z)\,dz -\int_{\gamma_4}f(z)\,dz\\
  &=& \int_{\epsilon}^Rf(x)\,dx +\int_0^{\pi} f(Re^{it})iRe^{it}\,dt
    \int_{-R}^{-\epsilon}f(x)\,dx -\int_0^{\pi} f(\epsilon e^{it})i\epsilon
    e^{it}\,dt
  \end{eqnarray*}

If $Q = R$ or $\epsilon$, then
  \[ \Big|\int_0^{\pi} f(Qe^{it})iQe^{it}\,dt\Big| \le
  \int_0^{\pi} |f(Qe^{it})|Q\,dt \le \frac{|\log Q| + \pi}{Q^4-1}Q\pi
  \longrightarrow 0 \]
as $Q = R \to \infty$ or $Q = \epsilon \to 0$.
Therefore, letting $R\to \infty$ and $\epsilon\to0$, we get
\begin{eqnarray*}
\int_\gamma f(z)\,dz  &=& \int_0^\infty f(x)\,dx + \int_{-\infty}^0 f(x)\,dx
= \int_0^\infty\frac{\log(x)}{x^4+1}\,dx
+ \int_{-\infty}^0\frac{\log|x| + i\pi}{x^4+1}\,dx\\
&=& \int_0^\infty\frac{\log(x)}{x^4+1}\,dx
+ \int_0^{\infty}\frac{\log(x) + i\pi}{x^4+1}\,dx
= 2\int_0^\infty\frac{\log(x)}{x^4+1}\,dx
+ i\pi \int_0^{\infty}\frac{1}{x^4+1}\,dx\\
\end{eqnarray*}
Comparing real and imaginary parts with (*), we conclude that
  \[ \int_0^\infty\frac{\log(x)}{x^4+1}\,dx = -\frac{\pi^2}{8\sqrt{2}}
  \quad\mbox{and}\quad
  \int_0^{\infty}\!\!\!\frac{1}{x^4+1}\,dx = \frac{\pi}{2\sqrt{2}} \]

\end{enumerate}

\end{document}
