\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\p}{\partial}
\parindent=0pt
\begin{document}

{\bf Question}

\begin{itemize}
\item[a)]
Show that the function $f(z)=\bar{z}^2z$ is differentiable only at
$z=0$.


\item[b)]
Show that $u=\sin x\sinh y$ is a harmonic function and find a
conjugate harmonic function $v$.  Express the regular function
$u+iv$ in terms of $z=x+iy$.


\item[c)]
Evaluate $\ds\int_{\gamma}\log zdz$ where $\gamma$ is the upper
half of the unit circle from $z=1$ to $z=-1$ and the branch of
$\log z$ is chosen which is 0 at $z=1$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$f(z)=\bar{z}^2z=(x-iy)^2(x+iy)=x^3+xy^2+i(-x^2y-y^3)=u+iv$

$\ds\frac{\p u}{\p x}=3x^2+y^2 \hspace{0.3in} \frac{\p v}{\p
y}=x^2-3y^2 \hspace{0.3in} \frac{\p u}{\p y}=2xy \hspace{0.3in}
\frac{\p v}{\p x}=-2xy$

$\ds\frac{\p u}{\p y}=-\frac{\p v}{\p x} \hspace{0.3in} \frac{\p
u}{\p x}=\frac{\p v}{\p y}$ iff $3x^2+y^2=-x^2-3y^2$

i.e. $4x^2=-4y^2 \hspace{0.3in}$ i.e $x=y=0$

So the Cauchy-Riemann equations are satisfied only at $z=0$.  The
partial derivatives are continuous there so $f(z)$ is
differentiable at $z=0$.


\item[b)]
$u=\sin x\sinh y$

$\ds\frac{\p u}{\p x}=\cos x\sinh y \hspace{0.3in} \frac{\p^2u}{\p
x^2}=-\sin x\sinh y$

$\ds\frac{\p u}{\p y}=\sin x\cosh y \hspace{0.3in} \frac{\p^2u}{\p
y^2}=\sin x\sinh y$

So $\ds\frac{\p^2u}{\p x^2}+\frac{\p^2u}{\p y^2}=0 \hspace{0.3in}$
i.e $u$ is harmonic.

$\ds\frac{\p u}{\p x}=\cos x\sinh y=\frac{\p v}{\p y}$

So $v=\cos x\cosh y+\phi(x)$

$\ds-\frac{\p u}{\p y}=-\sin x\cosh y=\frac{\p v}{\p x}$

So $v=\cos x\cosh y+\psi(y)$

Therefore $v=\cos x\cosh y+k$

Take $k=0$,

$u+iv=\sin x\sinh y+i\cos x\cosh y$

$\hspace{0.2in}=i(\cos x\cos iy-\sin x\sin iy)=i\cos(x+iy)=i\cos
z$


\item[c)]
$\log z=\log|z|+i\arg z$, so for $z=e^{it}$,

$\log z=\log1+it=it$

So $\ds\int_C\log zdz=\int_0^\pi itie^{it}dt=-\int_0^\pi
te^{it}dt$

$\ds\hspace{0.3in}=i\left[\frac{te^{it}}{i}\right]_0^\pi+
\frac{1}{i}\int_0^\pi e^{it}dt=-\frac{\pi e^{i\pi}}{i}+
\left[-e^{it}\right]_0^\pi$

$\ds\hspace{0.3in}=-\pi i+1+1=2-\pi i$

\end{itemize}

\end{document}