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{\bf Question}

State, without proof, the Cauchy-Riemann equations giving a
necessary condition for a complex function $f(z)=u(x,y)+iv(x,y)$
to be analytic in a region $A$ and state sufficient conditions,
involving the Cauchy-Riemann equations, for $f$ to be analytic in
$A$.

Show that in polar co-ordinates the Cauchy-Riemann equations
become

$$r\frac{\p u}{\p r}=\frac{\p v}{\p\theta}, \hspace{0.2in}
r\frac{\p v}{\p r}=-\frac{\p u}{\p\theta}.$$

Deduce that

$$(\cos(\log r)+i\sin(\log r))e^{-\theta}$$

defines an analytic function in some neighbourhood of each point
other than the origin.  Write this expression in the form $F(z)$,
where $z=re^{i\theta}$, and discuss the multi-valued nature of
$F$.

Choosing the branch of $F$ for which $F(1)=1$ compute
$\ds\int_{\delta}F(z)dz$, where $\delta$ is the upper half of the
unit circle from $z=1$ to $z=-1$.


\vspace{0.25in}

{\bf Answer}

The Cauchy-Riemann equations are $\ds \frac{\p u}{\p x}=\frac{\p
v}{\p y}, \hspace{0.2in} \frac{\p v}{\p x}=-\frac{\p u}{\p y}$
which must hold at each point of $A$.  For sufficiency we need the
additional conditions that the partial derivatives should be
continuous in $A$.

Using the chain rule gives, with $x=r\cos\theta \,\,\,
y=r\sin\theta$

\begin{eqnarray*}\frac{\p u}{\p r} &=& \frac{\p u}{\p
x}\cos\theta + \frac{\p u}{\p y}\sin\theta\\ \frac{\p u}{\p\theta}
&=& -\frac{\p u}{\p x}r\sin\theta + \frac{\p u}{\p y}r\cos\theta\\
\frac{\p v}{\p r} &=& \frac{\p v}{\p x}\cos\theta + \frac{\p v}{\p
y}\sin\theta\\ \frac{\p v}{\p\theta} &=& -\frac{\p v}{\p
x}r\sin\theta + \frac{\p v}{\p y}r\cos\theta\end{eqnarray*}

\newpage

So \begin{eqnarray*} r\frac{\p u}{\p r} &=& \frac{\p u}{\p
x}r\cos\theta+\frac{\p u}{\p y}r\sin\theta\\ &=&\frac{\p v}{\p
y}r\cos\theta-\frac{\p v}{\p x}r\sin\theta=\frac{\p
v}{\p\theta}\end{eqnarray*}

Also \begin{eqnarray*} r\frac{\p v}{\p r} &=& \frac{\p v}{\p
x}r\cos\theta+\frac{\p v}{\p y}r\sin\theta\\ &=& -\frac{\p v}{\p
y}r\cos\theta+\frac{\p u}{\p x}r\sin\theta=-\frac{\p
u}{\p\theta}\end{eqnarray*}

${}$

$u=\cos(\log r)e^{-\theta} \hspace{0.5in} v=\sin(\log
r)e^{-\theta}$

\begin{eqnarray*} \frac{\p u}{\p r} &=& -\sin(\log
r)\frac{1}{r}e^{-\theta}\\ \frac{\p v}{\p\theta} &=& -\sin(\log
r)e^{-\theta} {\rm \ so \ } r\frac{\p u}{\p r}=\frac{\p
v}{\p\theta}\\ \frac{\p v}{\p r} &=& \cos(\log
r)\frac{1}{r}e^{-\theta}\\ \frac{\p u}{\p\theta} &=& -\cos(\log
r)e^{-\theta}{\rm \ so \ } r\frac{\p v}{\p r}=-\frac{\p
u}{\p\theta}\end{eqnarray*}

The partial derivatives are continuous except at $z=0$, so the $u$
and $v$ define an analytic function except at $z=0$.

$(\cos(\log r)+i\sin(\log r))e^{-\theta}=e^{i\log
r}e^{-\theta}=e^{i(\log r+i\theta)}=e^{i\log z}=z^i$

$z^i$ is multi-valued because replacing $\log r$ by $\log r+2n\pi$
gives the same answer.

On the unit circle $r=1$, so with $F(z)=e^{-\theta}$ we have
$F(1)=e^{-0}=1$.

$\ds\int_{\delta}F(z)dz=\int_0^{\pi}e^{-\theta}ie^{i\theta}d\theta$

$\ds=\int_0^\pi ie^{i\theta(1+i)}d\theta=
\left[\frac{e^{i\theta(1+i)}}{1+i}\right]_0^\pi=
\frac{e^{i\pi(1+i)}-1}{1+i}=\frac{-e^{-\pi}-1}{1+i}$

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