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\title{Math 122B: Midterm}
\date{}


\theoremstyle{definition} 
\newtheorem{definition}{Definition}[section]
\newtheorem{conjecture}{Conjecture}[section]
\newtheorem{example}{Example}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{remark}{Remark}[section]
\newtheorem{question}{Question}[section]
\newtheorem{proposition}{Proposition}[section]
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem*{notation}{Notation}
\newtheorem{problem}{Problem}[section]

\begin{document}
\maketitle
\begin{enumerate}
\item Show that if $f$ is holomorphic in $D$, then
\begin{equation*}
f'(z_0) = \frac{d f}{dz}(z_0) = \frac{\dd f}{\dd x}(z_0) = -i\frac{\dd f}{\dd y}(z_0)
\end{equation*}
\item Compute the integral $\displaystyle \int_{-\infty}^\infty \frac{\cos(2x)dx}{x^4 + 1}$.  Make sure to show that the answer you obtain is a real number.
\item Compute $\displaystyle \int_{|z|=2}\frac{dz}{(z^{2016}+1)(z-3)(z-4)}$.
\item Show that if $f$ is holomorphic on $\mathbb{C}$ and $|f(z)| \leq |z|$ for all $z\in \mathbb{C}$, then $f$ is a linear polynomial.
\item Compute the Laurent series of $\displaystyle e^{z+\frac{1}{z}}$.  Then compute the integral $\displaystyle \int_{|z|=1}e^{z+\frac{1}{z}}$.  Show that your answer is a finite number.
\end{enumerate}

\section*{Solutions}
\begin{enumerate}
\item Let $z_0 = x_0+iy_0$ and $f = u + iv$ be holomorphic. Then the derivative by definition is
\begin{equation*}
\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} = \frac{df}{dz}(z_0)
\end{equation*}
which exists for any path $h \to 0$ and they are equal.  Consider $h\in \mathbb{R}$, then
\begin{equation*}
\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h} = \lim_{h\to 0}\frac{f(x_0+h,y_0)-f(z_0)}{h} = \frac{\dd f}{\dd x}(z_0)
\end{equation*}
on the other hand,
\begin{equation*}
\lim_{h\to 0}\frac{f(z_0+ih)-f(z_0)}{ih} = \lim_{h\to 0}\frac{f(x_0,y_0+h)-f(z_0)}{ih} = -i\frac{\dd f}{\dd y}(z_0).
\end{equation*}

\item We use the semicircle contour $C_R$, with $\Gamma_R$ denoting the upper half circle.  Then
\begin{align*}
\int_{C_R} \frac{e^{i2z}dz}{z^4+1} = \int_{-R}^R \frac{e^{i2z}dz}{z^4+1} + \int_{\Gamma_R} \frac{e^{i2z}dz}{z^4+1}.
\end{align*}
By residue theorem, the integral over the closed contour is
\begin{equation*}
2\pi i(\Res( \frac{e^{i2z}}{z^4+1}, e^{i\pi/4}) + \Res( \frac{e^{i2z}}{z^4+1}, e^{3\pi/4}))
\end{equation*}
Computing, we have
\begin{align*}
\Res( \frac{e^{i2z}}{z^4+1}, e^{i\pi/4}) &= \frac{e^{-\sqrt{2}+i\sqrt{2}}}{4e^{3i\pi/4}} \\
\Res( \frac{e^{i2z}}{z^4+1}, e^{3\pi/4}) &= \frac{e^{-\sqrt{2}-i\sqrt{2}}}{4e^{9i\pi/4}}.
\end{align*}

We also have
\begin{equation*}
\left|\int_{\Gamma_R} \frac{e^{i2z}dz}{z^4+1}\right| \leq \frac{\pi R}{R^4+1} \to 0
\end{equation*}
as $R\to \infty$.

Since $\displaystyle \Real\left( \int_{-R}^R\frac{e^{i2z}}{z^4+1}dz \right) = \int_{-R}^R \frac{\cos(2z)dz}{z^4+1}$, taking the real part of the sum of the residue gives us the answer.
\item By residue theorem, the singularity inside $|z|=2$ are the 2016-th roots of unity, denoted $\alpha_n$, hence
\begin{equation*}
\int_{|z|=2}\frac{dz}{(z^{2016}+1)(z-3)(z-4)} = 2\pi i\sum_{n=1}^{2016}\Res(\frac{1}{(z^{2016}+1)(z-3)(z-4)},\alpha_n).
\end{equation*}
To find a closed form, we use the fact that the sum of the residues is equal to zero hence
\begin{align*}
\sum_{n=1}^{2016}\Res(f,\alpha_n)= -\Res(f,3)-\Res(f,4)-\Res(f,\infty),
\end{align*}
where $\displaystyle f=\frac{1}{(z^{2016}+1)(z-3)(z-4)}$. Since
\begin{align*}
\Res(f,3) &= -\frac{1}{2016(3^{2015})}\\
\Res(f,4) &= \frac{1}{2016(4^{2015})}\\
\Res(f,\infty) &= 0, 
\end{align*}
we get
\begin{equation*}
\int_{|z|=2}\frac{dz}{(z^{2016}+1)(z-3)(z-4)} = 2\pi i\left(\frac{1}{2016(3^{2015})} - \frac{1}{2016(4^{2015})}\right).
\end{equation*}
\item Let $z\in \mathbb{R}$.  by Cauchy integral formula for the derivative, we have for any $R>0$,
\begin{equation*}
f'(z) = \frac{1}{2\pi i}\int_{|w-z|=R}\frac{f(w)}{(w-z)^2}dw.
\end{equation*}
Now
\begin{align*}
|f'(z)| &\leq \frac{1}{2\pi}\int_{|w-z|=R}\frac{|f(w)|}{|w-z|^2}dw \\
&\leq \frac{1}{2\pi R^2}(R+|z|) \int_{|w-z|=R}dw \\
&\leq \frac{R+|z|}{R} \leq 2,
\end{align*}
for $R>|z|$, here we use the inequality $|w| = |w-z + z| \leq |w-z|+|z|$ in the numerator.  Since $f'$ is bounded and holomorphic in $\mathbb{C}$, $f'\equiv A$, where $A$ is a constant.  This implies that $f(z) = Az+B$ for some constant $B$, however, $f(0) = 0$ hence $B=0$.  Therefore $f(z) = Az$.
\item To compute the Laurent series, we compute the simpler terms and take their product.  We have
\begin{align*}
e^z &= \sum_{n=0}^\infty \frac{z^n}{n!}\\
e^{\frac{1}{z}} &=\sum_{n=0}^\infty \frac{1}{n!z^n}.
\end{align*}
Taking their product, we have
\begin{equation*}
\sum_{n=0}^\infty \sum_{k=0}^\infty \frac{z^{n-k}}{n!k!} = \sum_{p=-\infty}^\infty\left( \sum_{n-k=p}\frac{1}{n!k!}\right)z^p.
\end{equation*}
Since there is an isolated singularity at $z=0$, the residue at 0 is given by the $z^{-1}$ coefficient, namely
\begin{equation*}
\Res(e^{z+\frac{1}{z}}, 0) = \sum_{n-k=-1}\frac{1}{n!k!} = \sum_{n=0}^\infty\frac{1}{n!(n+1)!}
\end{equation*}
hence by the residue theorem,
\begin{equation*}
\int_{|z|=1}e^{z+\frac{1}{z}}dz = 2\pi i \sum_{n=0}^\infty\frac{1}{n!(n+1)!}.
\end{equation*}
By comparison test, we have
\begin{equation*}
\sum_{n=0}^\infty \frac{1}{n!(n+1)!} \leq \sum_{n=1}^\infty \frac{1}{n^2} < \infty
\end{equation*}
\end{enumerate}

\end{document}