- #1
BrainHurts
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Homework Statement
Define [itex] f : \mathbb{C} \rightarrow \mathbb{C} [/itex] by
[itex] f(z) = \left
\{
\begin{array}{11}
|z|^2 \sin (\frac{1}{|z|}), \mbox{when $z \ne 0$}, \\
0, \mbox{when z = 0} .
\end{array}
\right.
[/itex]
Show that f is complex-differentiable at the origin although the partial derivative [itex]u_x[/itex] is not continuous at origin.
Homework Equations
The Attempt at a Solution
To show that [itex]f[/itex] is complex differentiable by defintion? In other words
[itex]f'(0)[/itex] = [itex] \lim_{h \rightarrow 0} [/itex] [itex] \frac{f(0+h) - f(0)}{h} [/itex] = [itex] \lim_{h \rightarrow 0} \frac{f(h) - 0}{h} [/itex] = [itex] \lim_{h \rightarrow 0} [/itex] [itex] \frac{|h|^2 \sin(\frac{1}{|h|})}{h} [/itex] = [itex] \lim_{h \rightarrow 0} [/itex] [itex] \frac{h\bar{h} \sin(\frac{1}{|h|})}{h} [/itex] = [itex] 0 [/itex] ?
Or am I missing something with [itex] \bar{h} [/itex]. Because I'm assuming as h approaches 0, so does [itex]\bar{h}[/itex]
Also, I see that
[itex]u_x(0,0) [/itex] = [itex] \lim_{x \rightarrow 0} [/itex] [itex] \frac{u(x,0)}{x}[/itex]
A little help here, not sure how to approach this problem.
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