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  \textbf{\color{darkblue}MAT 103A: Complex Analysis (W21)}\quad
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  \fbox{\parbox{5.5in}{
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    \textbf{\large  Your Name: } 
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 \noindent Please review the problem set guidelines in the \href{https://people.ucsc.edu/~jbreland/teaching/W21_MATH103A/build/W21_103A_syllabus.pdf}{syllabus} before turning in your work. 
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\item Compute the following limits and prove your claim by using only the $\e-\delta$ definition. 
\begin{enumerate}[(a)]
	\item $\ds \lim_{z \to i} \conj{z}$
	
	\item $\ds \lim_{z\to 1+i}z^2$. 
\end{enumerate}





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\item Let $M(z) = \frac{z-3}{1-2z}$. Prove that $\ds \lim_{z \to \infty} M(z) = -1/2$ and $\ds \lim_{z\to 1/2} M(z) = \infty$.


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\item Consider the complex valued function 
\[
f(z) = \begin{cases}
\ds \frac{\conj{z}^2}{z}, & z\neq 0\\
0, & z = 0.
\end{cases}
\]
Show that $f$ is not differentiable at the origin.\footnote{In the lecture, we showed that $f$ satisfies the Cauchy-Riemann equations at the origin. Thus, this will show that the satisfaction of the Cauchy-Riemann equations is not sufficient for the differentiability of $f$.}


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\item Let $U$ be a domain and $f:U \to \C$ a complex valued function that is differentiable at every point in $U$. Consider the domain $U^* \defeq \{ z \in \C : \conj{z} \in U\}$ and the function $f^*:U^* \to \C$ defined by $f^*(z) = \conj{f(\conj{z})}$. Show that $f^*$ is differentiable at every point in $U^*$. 


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\item For each function, determine all points at which the derivative exists. When the derivative exists, find its value. You must justify your claims. 

\begin{enumerate}[(a)]
	\item $f(z) = z+i\conj{z}$;
	
	\item $g(z) =  (z+i\conj{z})^2$;
	
	\item $h(z) = z\Im z$. 
\end{enumerate}

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\item \textbf{(Optional.})\footnote{This is worth 3 points extra credit, added to your score on this assignment.} By definition, a function $f:U\to \C$ is differentiable at $z_0 \in U$ if the limit $f'(z_0) = \lim_{z\to z_0} \frac{f(z) - f(z_0)}{z-z_0}$ exists. Unpacking the limit definition, we see that $f$ is differentiable at $z_0$ if and only if for all $\e >0$, there exists $\delta >0$ such that 
\[
0 < \abs{z-z_0} < \delta \text{  implies } \ds \abs{\frac{f(z)-f(z_0)}{z-z_0} - {\color{blue}f'(z_0)}} < \e.
\]
By appealing only to the definition, show that $f:\C \to \C$ defined by $f(z) = \conj{z}$ is not differentiable anywhere by completing the following steps:
\begin{enumerate}[(a)]
	\item Let $z_0 \in \C$ and assume that $f'(z_0)$ exists. Choose $\delta>0$ according to the definition using $\e = \frac{1}{2}$. 
	
	\item Consider $z = z_0 + \frac{\delta}{2}$ and conclude from (a) that $\abs{1-f'(z_0)} < \e$. 
	
	\item Consider $z = z_0 + \frac{i\delta}{2}$ and conclude from (a) that $\abs{1+f'(z_0)}<\e$. 
	
	\item Using the triangle inequality together with (b) and (c), obtain a contradiction. 
\end{enumerate}
\end{enumerate}

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