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

{\bf Question}

\begin{itemize}
\item[a)]
The function $f(z)$ of the complex variable $z=x+iy$ is
differentiable in a domain $D$.  Use the Cauchy-Riemann equations
to prove that for $z$ in $D$,

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

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


\item[b)]
Verify that the function $u(x,y)=e^y\cos x$ is harmonic.  find a
function $v(x,y)$ so that $f=u+iv$ is a differentiable function of
the complex variable $z=x+iy$.

Write $f$ as an explicit formula in $z$.

\end{itemize}

\vspace{0.25in}

{\bf Answer}

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

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

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

squaring and adding gives

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

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

so provided $f(z)\not=0$, dividing by $|f|^2$ gives the result.


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

So $\ds\frac{\p u}{\p x}=-e^y\sin x$ and $\ds\frac{\p^2u}{\p
x^2}=-e^y\cos x$

$\ds\hspace{0.2in}\frac{\p u}{\p y}=e^y\cos x\hspace{0.15in}$ and
$\ds\frac{\p^2u}{\p y^2}=e^y\cos x$

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

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

$\ds\hspace{0.3in}\frac{\p^2u}{\p y^2}=e^y\cos x=-\frac{\p v}{\p
x}$, so $\ds v=-e^y\sin x+\psi(y)$

Thus $\ds v=-e^y\sin x+C$ (constant)

Now $f=u+iv=e^y(\cos x-i\sin x)+k$

$\ds\hspace{0.5in}=e^ye^{-ix}+k=e^{-i(x+iy)}+k=e^{-iz}+k$

\end{itemize}

\end{document}