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\begin{coverpages}
\begin{center}{\huge SpTp: MATH 1199 - EXAM 1 FALL 2019 \\
%\vspace*{0.4in}
%\large Name: \makebox[3in]{\hrulefill}\\
SOLUTION
}
\end{center}
\begin{flushleft}\vspace{0.5in}
Wednesday, 18 September \\
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}
\iffalse
\begin{center}\underline{Formulas} \end{center}
\begin{itemize} 
\item \textbf{Cauchy-Riemann equations}: if $f=u+iv$, then the CR equations are $\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}$ \\

\item \textbf{Corollary to CR theorem}: If $f=u+iv$ and either $\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$ or $\dfrac{\partial u}{\partial y}=-\dfrac{\partial v}{\partial x}$ fail to hold at $(x_0,y_0)$, then $f'(x_0+iy_0)$ does not exist. \\

\item ``$\displaystyle\lim_{z \rightarrow z_0} f(z)=L$" means ``$\forall \epsilon>0 \exists \delta > 0 ( \text{if } |z-z_0|<\delta, \text{ then } |f(z)-f(z_0)|<\epsilon)$" \\

\item \textbf{Euler's formula}: if $\theta \in \mathbb{R}$, then $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ 
\item $f'(z) = \displaystyle\lim_{h \rightarrow 0} \dfrac{f(z+h)-f(z)}{h}$ 

\item ``$f$ is continuous at $z_0$" means ``$\displaystyle\lim_{z \rightarrow z_0} f(z) = f(z_0)$"
\item \textbf{Limit theorem with $\infty$}: If $z_0, w_0 \in \mathbb{C}$, then 
\begin{enumerate}[(i)]
\item $\displaystyle\lim_{z \rightarrow z_0} f(z) = \infty$ if and only if $\displaystyle\lim_{z \rightarrow z_0} \dfrac{1}{f(z)}=0$,
\item $\displaystyle\lim_{z \rightarrow \infty} f(z)=w_0$ if and only if $\displaystyle\lim_{z \rightarrow 0} f \left( \dfrac{1}{z} \right) = w_0$, and
\item $\displaystyle\lim_{z \rightarrow \infty} f(z)=\infty$ if and only if $\displaystyle\lim_{z \rightarrow 0} \dfrac{1}{f\left(\frac{1}{z} \right)} = 0$
\end{enumerate}

\item $-\pi < \mathrm{Arg}(z) \leq \pi, \quad \mathrm{arg}(z)=\mathrm{Arg}(z)+2n\pi, \quad n \in \mathbb{Z}$
 \item $\mathrm{Log}(z)=\ln(|z|)+i\mathrm{Arg}(z)$ 
\item if $(x,y)$ is in QI or QIV, then its angle is given by $\arctan(\frac{y}{x})$ \\
if $(x,y)$ is in QII or QIII, then its angle is given by $\pi+\arctan(\frac{y}{x})$
\end{itemize}
\fi
\end{coverpages}

\begin{questions}
\question (15 points) \\
\begin{parts}
\part[3] Write the Cauchy-Riemann equations for a function $f=u+iv$. \\
\textit{Solution}: $\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y}$ and $\dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}$
\part[3] Define the notation $\displaystyle\lim_{z \rightarrow z_0} f(z)=L$. \\
\textit{Solution}: For every $\epsilon > 0$ there exists $\delta>0$ such that if $0<|z-z_0|<\delta$, then $|f(z)-L|<\epsilon$.
\part[3] Define the derivative $f'(z)$ in terms of a limit. \\
\textit{Solution}: $f'(z) = \displaystyle\lim_{h \rightarrow 0} \dfrac{f(z+h)-f(z)}{h}$
\part[3] Define what it means for a function $f \colon U \rightarrow \mathbb{C}$ to be continuous at $z_0 \in U$. \\
\textit{Solution}: $\displaystyle\lim_{z \rightarrow z_0} f(z) = f(z_0)$
\part[3] Explain the difference between $\mathrm{arg}$ and $\mathrm{Arg}$. \\
\textit{Solution}: $\mathrm{Arg}(z)$ is the angle of the point $z \in \mathbb{C}$ with the restriction that $-\pi < \mathrm{Arg}(z) \leq \pi$ while $\mathrm{arg}(z)$ is multivalued and obeys $\mathrm{arg}(z)=\mathrm{Arg}(z) + 2n\pi$, for $n \in \mathbb{Z}$
\end{parts}

\question[15] Write in the form $a+bi$ for $a, b \in \mathbb{R}$.
\begin{parts}
\part[3] $(3-i)(2+i)$ \\
\textit{Solution}: Compute
\[ (3-i)(2+i) = 6+3i-2i-i^2 = 6+i+1 = 7+i.\]
\part[3] $\dfrac{1}{1-i}$ \\
\textit{Solution}: Compute
\[\dfrac{1}{1-i} = \dfrac{1}{1-i} \left( \dfrac{1+i}{1+i} \right) = \dfrac{1+i}{(1-i)(1+i)} = \dfrac{1+i}{1-i^2} = \dfrac{1}{2} + \dfrac{1}{2}i. \]
\part[3] $\overline{3+i}-2i$ \\
\textit{Solution}:  Compute
\[ \overline{3+i}-2i = (3-i)-2i = 3-3i.\]
\part[3] $\mathrm{Log}(-1+i)$ \\
\textit{Solution}: Since $|-1+i|=\sqrt{(-1)^2+1^2}=\sqrt{2}$ and $\mathrm{Arg}(-1+i)=\dfrac{3\pi}{4}$, it has polar form $-1+i=\sqrt{2}e^{\frac{\pi}{4}i}$. Therefore,
\[ \mathrm{Log}(-1+i) = \ln\left(\sqrt{2} \right) + \dfrac{\pi}{4}i = \dfrac{1}{2}\ln(2) + \dfrac{\pi}{4}i.\]
\part[3] $5e^{-\frac{\pi}{4}i}$ \\
\textit{Solution}: Using the Euler formula,
\[ 5e^{-\frac{\pi}{4}i} = 5 \left( \cos \left( -\dfrac{\pi}{4} \right) +  i \sin \left( -\dfrac{\pi}{4} \right) \right) = 5 \left( \dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} i \right) = \dfrac{5\sqrt{2}}{2} - \dfrac{5\sqrt{2}}{2} i. \]
\end{parts}
\newpage 
\question[6] Sketch a picture of the set in the complex plane defined by the inequality $|z-2i|>1$. \\
\textit{Solution}: This inequality describes the interior of the ball of radius $1$ centered at the point $2i$. Therefore we draw it as
\begin{center} \begin{tikzpicture}
\draw[fill=gray!50, dashed] (0,2) circle (1);
\draw (0,-1) -- (0,3);
\draw (-3,0) -- (3,0); 
\draw[fill=black] (0,2) circle (0.05);
\draw (0.3,2) node {$2i$};
\draw[fill=black] (0,1) circle (0.05);
\draw (0.3,0.85) node {$i$};
\draw[fill=black] (0,3) circle (0.05);
\draw (0.3,3.15) node {$3i$};
\end{tikzpicture} \end{center}
\question[9] Put the number into polar form $re^{i\theta}$, where $r, \theta \in \mathbb{R}$.
\begin{parts}
\part[3] $-6$ \\
\textit{Solution}: Clearly, $|-6|=6$ and since it lies on the negative real axis, $\mathrm{Arg}(-6)=\pi$. Therefore,
\[ -6 = 6 e^{\pi i}.\]
\part[3] $13i$ \\
\textit{Solution}: Clearly $|13i|=13$ and since it lies on the positive imaginary axis, $\mathrm{Arg}(13i)=\dfrac{\pi}{2}$. Therefore,
\[ 13i = 13e^{\frac{\pi}{2}i}.\]
\part[3] $-1-i$ \\
\textit{Solution}: We observe $|-1-i|=\sqrt{2}$ and since $-1-i$ is in quadrant IV, an angle for it argument can be found by computing 
\[ \pi + \arctan \left( \dfrac{-1}{-1} \right) = \pi + \arctan(1) = \pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}.\]
(\textit{note: this angle is not $\mathrm{Arg}(-1-i)$, but that's okay for polar form}). Therefore,
\[ -1-i = \sqrt{2} e^{\frac{5\pi}{4}i}.\]
\end{parts}

\question[8] Find all $5$th roots of $-i$. \\
\textit{Solution}: First write $-i$ in polar form: $-i=e^{-\frac{\pi}{2}i}$. Now we may compute the $5$th roots of $-i$:
\[ \left( -i \right)^{\frac{1}{5}} = \left( e^{-\frac{\pi}{2}i} \right)^{\frac{1}{5}} = \exp \left( \left( -\dfrac{\pi}{10} + \dfrac{2\pi k}{5} \right)i \right) = \exp \left( \left( \dfrac{-\pi+4\pi k}{10} \right)i \right), \quad k \in \mathbb{Z}. \]
To find the roots, just plug in $5$ consecutive values of $k$:  \\
\begin{center}\begin{tabular}{l|l}
$k=$ & $(-i)^{\frac{1}{5}}=$ \\
\hline
$0$ & $\exp\left(-\frac{\pi}{10}i\right)$ \\
$1$ & $\exp \left(\frac{3\pi}{10}i\right)$ \\
$2$ & $\exp\left(\frac{7\pi}{10}i\right)$ \\
$3$ & $\exp\left(\frac{11\pi}{10}i\right)$ \\
$4$ & $\exp\left(\frac{15\pi}{10}i\right)$ 
\end{tabular}\end{center}

\question[8] Express $\left(\sqrt{3}+i \right)^8$ in the form $a+bi$ for $a,b \in \mathbb{R}$. \\
\textit{Solution}: Write $\sqrt{3}+i$ in polar form: first $|\sqrt{3}+i|=\sqrt{\left( \sqrt{3} \right)^2+1^2 } = \sqrt{4}=2$ and since it is in quadrant I, its angle can be obtained by $\arctan \left( \dfrac{1}{\sqrt{3}} \right) = \dfrac{\pi}{6}$. Therefore its polar form is $\sqrt{3}+i=2e^{\frac{\pi}{6}i}$. Therefore, we compute
\[ \left( \sqrt{3}+i \right)^8 = \left( 2e^{\frac{\pi}{6}i} \right)^8 = 2^8 e^{\frac{8\pi}{6}i} = 2^8 e^{\frac{4\pi}{3}i} = 2^8 \cos\left(\dfrac{4\pi}{3} \right) + 2^8 \sin \left( \dfrac{4\pi}{3} \right)i = -2^7 - 2^7 \sqrt{3}i. \]
\question[5] Explain what ``$\displaystyle\lim_{z \rightarrow \infty} \dfrac{z^2+i}{z-1}=\infty$" means. (\textit{do \textbf{not} compute or prove it, just explain the meaning of the notation}) \\
\textit{Solution}: It means that 
\[ \displaystyle\lim_{z \rightarrow 0} \dfrac{1}{\dfrac{\dfrac{1}{z^2}+i}{\dfrac{1}{z}-1}}=0\]
\question[8] Write the function $\left\{ \begin{array}{ll} 
f \colon \mathbb{C} \rightarrow \mathbb{C} \\ 
f(z)=\dfrac{|z|+1}{\overline{z}}
\end{array} \right.$ in the form $f=u+iv$, where $u, v \colon \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$. \\
\textit{Solution}: Let $z=x+iy$ and calculate
\begin{multline*} f(x+iy) = \dfrac{|x+iy|+1}{\overline{x+iy}} = \dfrac{\sqrt{x^2+y^2}+1}{x-iy} \\
= \dfrac{\sqrt{x^2+y^2}+1}{x-iy} \left( \dfrac{x+iy}{x+iy} \right) = \underbrace{\dfrac{x\sqrt{x^2+y^2}+x}{x^2+y^2}}_{=u(x,y)} = \underbrace{\dfrac{y\sqrt{x^2+y^2}+y}{x^2+y^2}}_{=v(x,y)}i
\end{multline*}

\question[10] Write an $\epsilon$-$\delta$ proof that the function $\left\{ \begin{array}{ll}
f \colon \mathbb{C} \rightarrow \mathbb{C} \\
f(z)=7z-i
\end{array} \right.$ is continuous. \\
\textit{Proof}: Let $\epsilon > 0$ and choose $\delta=\dfrac{\epsilon}{7}$. If $|z-z_0|<\delta$, then compute
\[ |f(z)-f(z_0)|=\left|(7z-i) - (7z_0-i) \right| = |7||z-z_0| < 7 \delta = 7 \left( \dfrac{\epsilon}{7} \right) = \epsilon,\]
completing the proof. $\blacksquare$

\question[8] Verify that the Cauchy-Riemann equations hold for the function $\left\{ \begin{array}{ll}
f \colon \mathbb{C} \rightarrow \mathbb{C} \\
f(z)=z^2+z
\end{array} \right.$ \\
\textit{Solution}: First we must write $f$ in the form $f=u+iv$. To do that, let $z=x+iy$ and compute
\[ f(x+iy) = (x+iy)^2+(x+iy) = \Big(x^2+2xyi  + i^2y^2 \Big)+x+iy = \underbrace{x^2-y^2+x}_{=u(x,y)} + \underbrace{2xy+y}_{=v(x,y)}i. \]
Now compute 
\[ \dfrac{\partial u}{\partial x} = \dfrac{\partial}{\partial x} x^2-y^2+x = 2x+1, \]
\[ \dfrac{\partial u}{\partial y} = \dfrac{\partial}{\partial y} x^2-y^2+x = 2x+1 = -2y, \]
\[ \dfrac{\partial v}{\partial x} = \dfrac{\partial}{\partial x} 2xy+y = 2y, \]
and
\[ \dfrac{\partial v}{\partial y} = \dfrac{\partial}{\partial y} 2xy+y = 2x+1.\]
We observe from the calculations that the Cauchy-Riemann equations hold.
\question[8] Explain why the function $\left\{ \begin{array}{ll}
f \colon \mathbb{C} \rightarrow \mathbb{C} \\
f(z)=\overline{z}-z
\end{array} \right.$ is not complex-differentiable. \\
\textit{Solution}: First we will write it in the form $f=u+iv$. Let $z=x+iy$ and compute
\[ f(x+iy) = \overline{x+iy} - (x+iy) = x-iy - x-iy = \underbrace{0}_{=u(x,y)} + \underbrace{(-2y)}_{=v(x,y)}i.\]
Now compute
\[ \dfrac{\partial u}{\partial x} = \dfrac{\partial}{\partial x} 0 = 0,\]
and
\[ \dfrac{\partial v}{\partial y} = \dfrac{\partial}{\partial y} -2y = -2.\]
In this case, we see that $\dfrac{\partial u}{\partial x} \neq \dfrac{\partial v}{\partial y}$ and by the corollary to the Cauchy-Riemann equations theorem, we conclude that $f$ is not complex-differentiable. 
\end{questions}
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