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

{\bf Question}

\begin{itemize}
\item[a)]
Use the Cauchy-Riemann equations to prove that the regular
function $f(z)$, with $z=x+iy$, satisfies

$$\left(\frac{\p|f|}{\p x}\right)^2+\left(\frac{\p|f|}{\p
y}\right)^2=|f'(z)|^2$$

when $f(z)\not=0$.


\item[b)]
Find a regular function of $z=x+iy$ with real part $e^y\cos x$.

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[a)]
$|f|^2=u^2+v^2$

So $\ds2|f|\frac{\p|f|}{\p x}=2u\frac{\p u}{\p x}+2v\frac{\p v}{\p
x}$

and $\ds2|f|\frac{\p|f|}{\p y}=2u\frac{\p u}{\p y}+2v\frac{\p
v}{\p y}=-2u\frac{\p v}{\p x}+2v\frac{\p u}{\p x}$

$\ds|f|^2\left(\left(\frac{\p|f|}{\p
x}\right)^2+\left(\frac{\p|f|}{\p y}\right)^2\right)=
(u^2+v^2)\left(\frac{\p u}{\p x}\right)^2+(u^2+v^2)\left(\frac{\p
v}{\p x}\right)^2$

$\ds\hspace{0.2in}=|f|^2\left|\frac{\p u}{\p x}+i\frac{\p v}{\p
x}\right|^2=|f|^2|f'(z)|^2$

So $\ds\left(\frac{\p|f|}{\p x}\right)^2+\left(\frac{\p|f|}{\p
y}\right)^2=|f'(z)|^2$ if $|f|\not=0$


\item[b)]
$u=e^y\cos x$

$\ds\frac{\p u}{\p x}=-e^y\sin x=\frac{\p v}{\p y} \Rightarrow
v=-e^y\sin x+\phi(x)$

$\ds-\frac{\p u}{\p y}=-e^y\cos x=\frac{\p v}{\p x} \Rightarrow
v=-e^y\sin x+\psi(y)$

so $v=-e^y\sin x$ will do

$f=u+iv=e^y(\cos x-i\sin x)=e^{y-ix}=e^{i(x+iy)}=e^{iz}$


\end{itemize}

\end{document}