\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)]
Prove that $|\sin z|^2+|\cos z|^2\geq1$, with equality if and only
if $z$ is a real number.

\item[c)]
Evaluate the integral $\int_C\tan zdz$, where $C$ is the straight
line segment from $z=0$ to $z=\frac{1}{2}\pi(1+i)$.  Express your
answer in the form $a+ib$ where $a$ and $b$ are real.
\end{itemize}



\vspace{0.25in}

{\bf Answer}

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

Now $\ds\frac{\p u}{\p x}=3x^2+y^2 \hspace{0.5in} \frac{\p v}{\p
y}=-x^2-3y^2$

$\ds\hspace{0.35in}\frac{\p u}{\p y}=2xy \hspace{0.85in} \frac{\p
v}{\p x}=-2xy$

Now $\ds\frac{\p u}{\p y}\equiv-\frac{\p v}{\p x} \hspace{0.2in}$
but $\ds\frac{\p u}{\p x}=\frac{\p v}{\p y}$ iff
$3x^2+y^2=-x^2-3y^2$

iff $4x^2=-4y^2$, iff $x=y=0$.

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

\item[b)]
Using the various facts about trigonometric functions we have

$|\sin z|^2+|\cos z|^2=
\sin^2x\cosh^2y+\cos^2x\sinh^2y$

$\hspace{0.8in}+\cos^2x\cosh^2y+\sin^2x\sinh^2y$

$=\cosh^2y+\sinh^2y=\cosh2y\geq1$ with equality iff $y=0$ i.e. if
$z$ is real.

\item[c)]
$\ds\int_C\tan zdz=[-{\rm Log}(\cos z)]_0^{(1+i)\frac{\pi}{2}}$
(integral of a continuous derivative)

$\ds=-{\rm Log}(\cos(\frac{\pi}{2}+i\frac{\pi}{2}))= -{\rm
Log}(-\sin(i\frac{\pi}{2}))$

$\ds=-{\rm Log}(-i\sinh\frac{\pi}{2})=
-\ln(\sinh\frac{\pi}{2}+i\frac{\pi}{2}$

(Note: $\cos z=cos x\cosh y-i\sin x\sinh y$, so $Re(\cos z)\geq0$
along $C$. i.e. $-\frac{\pi}{2}<\arg z\frac{\pi}{2}$ along $C$,
hence we do not encounter singularities for ${\rm Log}(\cos z)$.
I do not expect this reasoning to appear in the students'
answers.)

\end{itemize}

\end{document}