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\begin{document}

{\bf Question}

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

$$f(z)=u(x,y)+iv(x,y), \hspace{0.5in} (z=x+iy)$$

to be differentiable as a function of a complex variable $z$.
State sufficient conditions involving the Cauchy-Riemann equations
for $f$ to be differentiable.

Hence find the points where the function

$$f(z)=\left\{\begin{array}{cc}\ds\frac{xy^4-ix^4y}{x^2+y^2} &
z\not=0\\ 0 & z=0\end{array}\right.$$

is differentiable.

At the points at which $f$ is differentiable calculate the
derivative $f'(z)$.


\vspace{0.25in}

{\bf Answer}

If $f$ is differentiable at $z=x+iy$,

\begin{eqnarray*}f'(z) &=&
\lim_{h\to0}\frac{u(x+h,y)+iv(x+h,y)-u(x,y)-iv(x,y)}{h}\\ &=&
\frac{\p u}{\p x}+i\frac{\p v}{\p x}\end{eqnarray*}

Also

\begin{eqnarray*}f'(z) &=&
\lim_{k\to0}\frac{u(x,y+k)+iv(x,y+k)-u(x,y)-iv(x,y)}{ik}\\ &=&
\frac{1}{i}\left(\frac{\p u}{\p y}+i\frac{\p v}{\p
y}\right)=\frac{\p v}{\p y}-i\frac{\p u}{\p y}\end{eqnarray*}

so $\ds\frac{\p u}{\p x}=\frac{\p v}{\p y}$ and $\ds\frac{\p v}{\p
x}=-\frac{\p u}{\p y}$

Sufficient conditions for differentiability at $z$ are that the
Cauchy-Riemann equations should be satisfied at $z$, and that the
partial derivatives should exist in a neighbourhood of $(x,y)$ and
be continuous at $(x,y)$.

${}$

Now $\ds u=\frac{xy^4}{x^2+y^2} \hspace{0.5in}
v=\frac{-x^4y}{x^2+y^2}$

\begin{eqnarray*}\frac{\p u}{\p x} &=& \frac{(x^2+y^2)y^4-xy^4(2x)}{(x^2+y^2)^2}=
\frac{y^6-x^2y^4}{(x^2+y^2)^2}\\ \frac{\p v}{\p y} &=&
\frac{(x^2+y^2)(-x^4)+x^4y(2y)}{(x^2+y^2)^2}=
\frac{y^2x^4-x^6}{(x^2+y^2)^2}\\ \frac{\p u}{\p y} &=&
\frac{(x^2+y^2)4xy^3-xy^4(2y)}{(x^2+y^2)^2}=
\frac{4x^3y^3+2xy^5}{(x^2+y^2)^2}\\ -\frac{\p v}{\p x} &=&
-\frac{(x^2+y^2)(-4x^3y)+x^4y(2x)}{(x^2+y^2)^2}=
\frac{4x^3y^3+2x^5y}{(x^2+y^2)^2}\end{eqnarray*}


For the Cauchy-Riemann equations to be satisfied we require

\begin{eqnarray} y^6-x^2y^4 &=& y^2x^4-x^6\\ 4x^3y^3+2xy^5 &=&
4x^3y^3+2x^5y\end{eqnarray}

From (2) $\hspace{0.3in} 2xy^5=2x^5y$

so for $xy\not=0 \hspace{0.3in} y^4=x^4 \hspace{0.3in}$ i.e.
$y=\pm x$

this also satisfies (1).

Now from (1) $\hspace{0.2in} x=0\Rightarrow y=0$ and $y=0
\Rightarrow x=0$

So if $y=\pm x\not=0$ the Cauchy-Riemann equations are satisfied
and the partial derivatives are continuous if $(x,y)\not=(0,0)$,
so $f$ is differentiable at $x(1\pm i)$ for $x\not=0$.

Now consider $z=0$.  Let $z=re^{i\theta}$

For $\ds r\not=0 \hspace{0.3in} f(re^{i\theta})=
\frac{(r^5\cos\theta\sin^4\theta-ir^5\cos^4\theta\sin\theta)}{r^2}$

so $\ds\left|\frac{f(re^{i\theta})-f(0)}{re^{i\theta}-0}\right|=
r^2|\cos\theta\sin^4\theta-i\cos^4\theta\sin\theta|$

$\ds\hspace{1.4in}\leq2r^2 \,\,\, \to0$ as $r\to0$

So $f'(0)=0$

$\ds f'(z)=\frac{\p u}{\p x}+i\frac{\p v}{\p
x}=\frac{y^2x^4-x^6}{(x^2+y^2)^2}-i\frac{4x^3y^3+2x^5y}{(x^2+y^2)^2}$

so when $\ds x=y \hspace{0.3in}
f'(z)=-i\frac{6x^6}{4x^4}=-i\frac{3}{2}x^2$

\hspace{0.1in} when $\ds x=-y \hspace{0.3in}
f'(z)=i\frac{6x^6}{4x^4}=i\frac{3}{2}x^2$

\end{document}
