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Complex Analysis, Complex Differentiable Question

  1. Aug 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Define [itex] f : \mathbb{C} \rightarrow \mathbb{C} [/itex] by

    [itex] f(z) = \left
    |z|^2 \sin (\frac{1}{|z|}), \mbox{when $z \ne 0$}, \\

    0, \mbox{when z = 0} .

    Show that f is complex-differentiable at the origin although the partial derivative [itex]u_x[/itex] is not continous at origin.

    2. Relevant equations

    3. 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.
    Last edited: Aug 28, 2013
  2. jcsd
  3. Aug 28, 2013 #2


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    Homework Helper

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