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QUESTION
(More difficult) Let
$f(z)=\left\{\begin{array}{rr}z^5/|z^4|,&z\neq0,\\0,&z=0\end{array}\right.$
and let $g(z)=\frac{f(z)}{z}$. Find $g(z)$ for $z$ on
the real axis and also for $z$ on the line $y=x$. Deduce that $f$
is is not differentiable at $z=0$. Now writing $f=u+iv$, where $u$
and $v$ are real, find $u(x,0),v(x,0), u(0,y),v(0,y)$ and show
that the Cauchy-Riemann equations hold at $z=0$. Comment.
ANSWER
If $z\not=0$, $g(z)={f(z)\over z}={z^4\over |z^4|}$. Hence on the
real axis $g(z)=1$. On the line $y=x$, $z=|z|e^{i\pi/4}$ so
${z^4\over |z^4|}=e^{i\pi}=-1$
Now $f^{'}(z)=lim_{h\rightarrow 0}{f(h)\over h}=lim_{h\rightarrow
0} g(h)$ which we have shown does not exist as it depends on the
direction that we approach 0. Thus $f(z)$ is not differentiable at
$z=0$. Now let $f(z)=u(x,y)+iv(x,y)=(x+iy)^5/(x^4+y^4)$. Putting
$y=0$, we get $u(x,0)+iv(x,0)=x$, so $u(x,0)=x$, $v(x,0)=0$.
Similarly, $u(0,y)=0$, $v(0,y)=y$. Thus $u_x=v_y=1$ and
$u_y=-v_x=0$, and the Cauchy-Riemann equations hold. There is no
contradiction as the Cauchy-Riemann equations do not {\it imply}
differentiability.
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