Complex Analysis, Complex Differentiable Question

[BrainHurts]

Homework Statement

Define $f : \mathbb{C} \rightarrow \mathbb{C}$ by

$f(z) = \left <br /> \{ <br /> \begin{array}{11} <br /> |z|^2 \sin (\frac{1}{|z|}), \mbox{when z \ne 0}, \\<br /> <br /> 0, \mbox{when z = 0} .<br /> \end{array}<br /> \right.$

Show that f is complex-differentiable at the origin although the partial derivative $u_x$ is not continuous at origin.

Homework Equations

The Attempt at a Solution

To show that $f$ is complex differentiable by definition? In other words

$f'(0)$ = $\lim_{h \rightarrow 0}$ $\frac{f(0+h) - f(0)}{h}$ = $\lim_{h \rightarrow 0} \frac{f(h) - 0}{h}$ = $\lim_{h \rightarrow 0}$ $\frac{|h|^2 \sin(\frac{1}{|h|})}{h}$ = $\lim_{h \rightarrow 0}$ $\frac{h\bar{h} \sin(\frac{1}{|h|})}{h}$ = $0$ ?

Or am I missing something with $\bar{h}$. Because I'm assuming as h approaches 0, so does $\bar{h}$

Also, I see that

$u_x(0,0)$ = $\lim_{x \rightarrow 0}$ $\frac{u(x,0)}{x}$

A little help here, not sure how to approach this problem.

 

[BruceW]

Or am I missing something with $\bar{h}$. Because I'm assuming as h approaches 0, so does $\bar{h}$

why not write out h in polar form? and see what that implies for $\bar{h}$ as h goes to zero