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\newcommand{\ds}{\displaystyle}
\newcommand{\p}{\partial}
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\begin{document}

{\bf Question}

Derive the Cauchy-Riemann equations as necessary conditions for
the function

$$f(z)=u(x,y)+iv(x,y), \, \, \, {\rm where\ \ \ } z=x+iy$$

to be analytic.  State without proof sufficient conditions for the
differentiability of $f$ in terms of the partial derivatives of
$u$ and $v$.

Let $\ds f(z)=\frac{z}{\bar{z}}, \hspace{0.5in} (z\not=0)$

Show that $f$ is differentiable nowhere in ${\bf C}-\{0\}$.  Is it
possible to extend the definition of $f$ to 0 in such a way that
$f$ is differentiable at 0?  Justify your answer.


\vspace{0.25in}

{\bf Answer}

Derivation of Cauchy-Riemann equations is bookwork.

${}$

Let $\ds f(z)=\frac{z}{\bar{z}}=\frac{x+iy}{x-iy}=
{(x+iy)^2}{x^2+y^2}=\frac{x^2-y^2}{x^2+y^2}+i\frac{2xy}{x^2+y^2}
\hspace{0.3in} z\not=0$

$\ds\frac{\p u}{\p x}=
\frac{(x^2+y^2)2x-(x^2-y^2)2x}{(x^2+y^2)^2}=
\frac{4xy^2}{(x^2+y^2)^2}$

$\ds\frac{\p v}{\p y}= \frac{(x^2+y^2)2y-2xy2x}{(x^2+y^2)^2}=
\frac{2y^3}{(x^2+y^2)^2}$

$\ds\frac{\p u}{\p x}=\frac{\p v}{\p y}$ iff $4xy^2=2y^3$ i.e.
$4x=2y$ or $y=0$

$\ds\frac{\p u}{\p y}=
\frac{(x^2+y^2)(-2y)-(x^2-y^2)2y}{(x^2+y^2)^2}=
\frac{-4x^2y}{(x^2+y^2)^2}$

$\ds\frac{\p v}{\p x}= \frac{(x^2+y^2)2x-2xy2y}{(x^2+y^2)^2}=
\frac{2x^3}{(x^2+y^2)^2}$

$\ds\frac{\p v}{\p x}=-\frac{\p u}{\p y}$ iff $4x^2y=2x^3$ i.e.
$4y=2x$ or $x=0$

The conditions cannot be satisfied for $z\not=0$.

Now in polars $z=re^{i\theta}$ and $\bar{z}=re^{-i\theta}$.  So
$\ds\frac{z}{\bar{z}}=e^{2i\theta}$

For $\theta=0$, $\ds\frac{z}{\bar{z}}=1$ and for
$\ds\theta=\frac{\pi}{2}$, $\ds\frac{z}{\bar{z}}=-1$

So $f(z)$ does not have a limit as $z\to0$ and so cannot be
defined to be differentiable or continuous at $z=0$.

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