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\begin{coverpages}
\begin{center}{\huge SpTp: MATH 1199 - EXAM 3 FALL 2019 \\
\vspace*{0.4in}
SOLUTION
%\large Name: \makebox[3in]{\hrulefill}\\
}
\end{center}

\begin{flushleft}\vspace{0.5in}
Friday, 22 November \\
Instructor: Tom Cuchta
\end{flushleft}

\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}

\begin{center}\underline{Formulas}\end{center}
\textbf{ML-inequality}: Let $C$ be a contour. If $f(z) \leq M$ on the contour $C$ and $L$ denotes the length of $C$, then 
\[ \left| \displaystyle\int_C f(z) \mathrm{d}z \right| \leq ML. \]
\textbf{Cauchy Integral Theorem}: If $C$ is a simple closed contour and $f \colon U \subset \mathbb{C} \rightarrow \mathbb{C}$ is complex-differentiable everywhere in the interior of $C$, then $\displaystyle\int_C f(z) \mathrm{d}z = 0$. \\[0.2in]
\textbf{Cauchy Integral Formula}: If $f \colon U \rightarrow \mathbb{C}$ is complex-differentiable on the interior of a simpled closed contour $C$, then for any $\zeta \in U$, 
\[ f(\zeta) = \dfrac{1}{2\pi i} \displaystyle\int_C \dfrac{f(z)}{z-\zeta} \mathrm{d}z.\]
If a function $f$ has a blow-up point at $z_0$ and a Laurent series of the form
\[ f(z) = \displaystyle\sum_{k=-\infty}^{\infty} c_k (z-z_0)^k,\]
then we say that the residue of $f$ at $z=z_0$ is the coefficient $c_{-1}$, i.e. we write
\[ \displaystyle\underset{z=z_0}{\mathrm{Res}} f(z) = c_{-1}.\]
\textbf{Cauchy Residue Theorem}: If $C$ is a contour and the poles $z_0,\ldots,z_n$ of a function $f$ are in the interior of $C$, then 
\[ \displaystyle\int_C f(z) \mathrm{d}z = 2\pi i \displaystyle\sum_{k=0}^n \underset{z=z_k}{\mathrm{Res}} f(z).\]
\textbf{Inversion of Laplace transform}: if $F$ has finitely many poles and the contour $C$ surrounds all of them, then we may compute the inverse Laplace transform of $F$ as
\[ \mathscr{L}^{-1}\{F\}(t) = \dfrac{1}{2\pi i} \displaystyle\int_C e^{zt}F(z) \mathrm{d}z \]
\end{coverpages}

\begin{questions}
\question[14] Consider the contour $C$ consisting of the upper half of the circle $|z|=3$. Use the ML inequality to bound the following: 
\[ \left| \displaystyle\int_C \dfrac{z-2}{z^2-7} \mathrm{d}z \right|.\]
\textit{Solution}: Bound the numerator with the triangle inequality:
\[ |z-2| \leq |z|+|2|=3+2=5,\]
and bound the denominator with the reverse triangle inequality:
\[ |z^2-7| \geq ||z|^2-|7|| = |3^2-7|=2,\]
hence
\[ \dfrac{1}{|z^2-7|} \leq \dfrac{1}{2}.\]
Therefore,
\[ \left| \dfrac{z-2}{z^2-7} \right| \leq \dfrac{5}{2} =: M.\]
The length of the semicircle $C$ is $L:=3\pi$. Therefore, by the ML-inequality
\[ \left| \displaystyle\int_C \dfrac{z-2}{z^2-7} \mathrm{d}z \right| \leq ML=\dfrac{15\pi}{2}.\]
\question[15] The Bessel function of order five, $J_5 \colon \mathbb{C} \rightarrow \mathbb{C}$, is a complex-differentiable function given by the infinite series $J_5(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^kz^{2k+5}}{k! (5+k)!2^{2k+5}}$. Find the residue at $z=0$ of the function $g(z)=z^4 J_5\left( \dfrac{1}{z} \right)$. \\
\textit{Solution}: Consider the series for $g$:
\[ \begin{array}{ll}
g(z)&=z^4 J_5\left(\dfrac{1}{7}\right) \\
&= z^4 \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(5+k)!2^{2k+5}z^{2k+5}} \\
&=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k}{k!(5+k)!2^{2k+5}z^{2k+1}} \\
&= \dfrac{(-1)^0}{0!5!2^{5}z} + \dfrac{(-1)^1}{1!6!2^7z^3} + \ldots
\end{array}\]
Since the residue is the coefficient of $\dfrac{1}{z}$, we conclude that
\[ \displaystyle\underset{z=0}{\mathrm{Res}} \hspace{2pt} g(z) = \dfrac{1}{5!2^5}.\]

\question[15] Use the inversion integral to find the inverse Laplace transform of $F(z)=\dfrac{1}{(z+2)^2}$ \\
\textit{Solution}: The poles of this function come from setting $(z+2)^2=0$, i.e. $z=-2$, a pole of order $2$. Therefore, for a closed contour $C$ with $-2$ interior to it,
\[ \begin{array}{ll}
\mathscr{L}^{-1}\{F\}(t) &= \dfrac{1}{2\pi i} \displaystyle\int_C e^{tz}F(z) \mathrm{d}z \\
&\stackrel{\text{res thm}}{=}\dfrac{1}{2\pi i} 2\pi i \displaystyle\underset{z=-2}{\mathrm{Res}} \hspace{2pt} \dfrac{e^{tz}}{(z+2)^2} \\
&\stackrel{\phi(z)=e^{zt}}{=}\dfrac{\phi'(-2)}{1!} \\
&\stackrel{\phi'(z)=te^{zt}}{=} te^{-2t}.
\end{array}\]
\vfill

\question[24] In this problem, you will find series, centered at zero, for the function $f(z)=\dfrac{1}{z-1} + \dfrac{1}{z-5}$. 
\begin{parts}
\part[8] Find a series for $f$ valid in the region $|z|<1$. \\
\textit{Solution}: For $|z|<1$, 
\[ (i) \hspace{35pt} \dfrac{1}{z-1} = -\dfrac{1}{1-z} = -\displaystyle\sum_{k=0}^{\infty} z^k, \quad \quad |z|<1,\]
and
\[ (ii) \hspace{35pt} \dfrac{1}{z-5} = -\dfrac{1}{5} \dfrac{1}{1-\frac{z}{5}} = -\dfrac{1}{5} \displaystyle\sum_{k=0}^{\infty} \left( \dfrac{z}{5} \right)^k = -\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{5^{k+1}}, \quad \quad \left|\frac{z}{5}\right|<1 \leftrightarrow |z|<5. \]
Therefore, for $|z|<1$, the following series is valid:
\[ f(z) = -\displaystyle\sum_{k=0}^{\infty} z^k - \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{5^{k+1}}.\]
\part[8] Find a series for $f$ valid in the region $1<|z|<5$. \\
\textit{Solution}: For $1<|z|<5$, the series (ii) is still valid, but the series (i) is no longer valid. Therefore,
\[ (iii) \hspace{35pt} \dfrac{1}{z-1} = \dfrac{1}{z} \dfrac{1}{1-\frac{1}{z}} = \dfrac{1}{z} \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{z^k} = \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{z^{k+1}}, \quad \quad \left| \frac{1}{z} \right|<1 \leftrightarrow |z|>1 \]
Therefore, for $1<|z|<5$, the following series is valid:
\[ f(z) = \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{z^{k+1}} - \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{5^{k+1}}.\]
\part[8] Find a series for $f$ valid in the region $|z|>5$. \\
\textit{Solution}: For $|z|>5$, the series (iii) is remains valid, but the series (ii) is no longer valid. Therefore,
\[ (iv) \hspace{35pt} \dfrac{1}{z-5} = \dfrac{1}{z} \dfrac{1}{1-\frac{5}{z}} = \dfrac{1}{z} \displaystyle\sum_{k=0}^{\infty} \dfrac{5^k}{z^k} = \displaystyle\sum_{k=0}^{\infty} \dfrac{5^k}{z^{k+1}}, \quad \quad \left| \frac{5}{z} \right|<1 \leftrightarrow |z|>5. \]
Therefore, for $|z|>5$, the following series is valid:
\[ f(z) = \displaystyle\sum_{k=0}^{\infty} \frac{1}{z^{k+1}} + \displaystyle\sum_{k=0}^{\infty} \dfrac{5^k}{z^{k+1}}. \]
\end{parts}
\newpage
\question[32] Contour integrals. For each part, \textit{\textbf{draw a sketch of all relevant blow-up points and the contour}} and calculate the integral in any way you wish.
\begin{parts}
\part[8] Let $C$ be the contour $|z|=3$ and calculate $\displaystyle\int_C \dfrac{1}{(z+2)(z+1)} \mathrm{d}z$. \\
\textit{Solution}: 
\begin{center}\begin{tikzpicture}
\draw (0,-3.2) -- (0,3.2);
\draw (-3.2,0) -- (3.2,0);
\node at (-2,0) {X};
\node at (-1,0) {X};
\node at (-2,-0.4) {$-2$};
\node at (-1,-0.4) {$-1$};
\draw (0,0) circle (3);
\end{tikzpicture}\end{center}
Both (order 1) poles appear interior to the contour, so we will use the residue theorem to compute the integral:
\[ \begin{array}{ll}
\displaystyle\int_C \dfrac{1}{(z+2)(z+1)} \mathrm{d}z &= 2\pi i \underset{z=-1}{\mathrm{Res}} \dfrac{\frac{1}{z+2}}{z-(-1)} + 2\pi i \underset{z=-2}{\mathrm{Res}} \dfrac{\frac{1}{z+1}}{z-(-2)} \\
&= 2\pi i \left( \dfrac{1}{-1+2} \right) + 2\pi i \left( \dfrac{1}{-2+1} \right) \\
&=2 \pi i \left( 1 + -1 \right) \\
&= 0.
\end{array} \]
\part[8] Let $C$ be the contour $|z-4|=1$ and calculate $\displaystyle\int_C \dfrac{1}{(z-2)(z+1)} \mathrm{d}z$. \\
\textit{Solution}:  \\
\begin{center}\begin{tikzpicture}
\draw (0,-3.2) -- (0,3.2);
\draw (-1.5,0) -- (5,0);
\node at (2,0) {X};
\node at (-1,0) {X};
\node at (4,-0.4) {$4$};
\node at (2,-0.4) {$2$};
\node at (-1,-0.4) {$-1$};
\draw (4,0) circle (1);
\end{tikzpicture}\end{center}
None of the poles appear interior to the contour. Therefore we may use the Cauchy integral theorem to say
\[ \displaystyle\int_C \dfrac{1}{(z-2)(z+1)}\mathrm{d}z = 0.\]
\part[8] Let $C$ be the contour $|z+2|=2$ and calculate $\displaystyle\int_C \dfrac{\sin(z)\cos(z)e^z}{z+1} \mathrm{d}z$. \\
\textit{Solution}: \\
\begin{center}\begin{tikzpicture}
\draw (0,-3.2) -- (0,3.2);
\draw (-5,0) -- (1,0);
\node at (-1,0) {X};
\node at (-2,-0.4) {$-2$};
\node at (-1,-0.4) {$-1$};
\draw (-2,0) circle (2);
\end{tikzpicture}\end{center}
The simple pole $-1$ appears interior to the contour. Therefore we use the residue theorem to compute
\[ \begin{array}{ll}
\displaystyle\int_C \dfrac{\sin(z)\cos(z)e^z}{z+1} \mathrm{d}z &= 2\pi i \underset{z=-1}{\mathrm{Res}} \dfrac{\sin(z)\cos(z)e^z}{z-(-1)} \\
&= 2\pi i \sin(-1)\cos(-1)e^{-1}.
\end{array} \]
\part[8] Let $C$ be the contour $|z|=1$ and calculate $\displaystyle\int_C \dfrac{e^z}{(z-2)(z+3)} \mathrm{d}z$.\\
\textit{Solution}: 
\begin{center}\begin{tikzpicture}
\draw (0,-3.2) -- (0,3.2);
\draw (-3.5,0) -- (2.5,0);
\node at (-3,0) {X};
\node at (2,0) {X};
\node at (2,-0.4) {$2$};
\node at (-3,-0.4) {$-3$};
\draw (0,0) circle (1);
\end{tikzpicture}\end{center}
None of the poles appear interior to the contour. Therefore we may use the Cauchy integral theorem to compute
\[ \displaystyle\int_C \dfrac{e^z}{(z-2)(z+3)} \mathrm{d}z = 0.\]
\end{parts}

\end{questions}

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