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Is the function $$f(z)=|z^{2}-z|$$ nowhere analytic? Justify your ansewr

WHat i tried

Let $z=x+iy$ and then substituting it to the function above to get the form $$f(x+iy)=u(x,y)+i(x,y)$$ where $u(x,y)$ is the real part and $v(x,y)$ the

imaginary part and then using the Cauchy Riemann equations to show that it holds.ie

$$U_{x}=V_{y}$$ $$U_{y}=-V_{x}$$ Then find the four partial derivatives $U_{x}$, $U_{y}$ , $V_{x}$, $V_{y}$ and if they are continuous then the function are analytic. My problem is that because of the modulus i couldent differentiate and find the partial derivatives in the usual way. Do i have to split up the modulus into the nehgative and positive portion and find the partial derivatives for each case. Could anyone explain. Thanks

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  • $\begingroup$ yes, do a case distinction $\endgroup$ – noctusraid Sep 5 '16 at 11:18
  • $\begingroup$ "nowhere analytic" means "no open set where the function is complex differentiable"? $\endgroup$ – zhw. Sep 5 '16 at 19:06
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Hint: The modulus of a complex number $a+ib$ is $\sqrt{a^2+b^2}$. Use this to identify $u$ and $v$ as functions of $x,y$. Hint2: There is one point at which $f$ is complex differentiable.

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    $\begingroup$ Wouldnt the imaginary portion of the complex number be gone if i do that ? $\endgroup$ – ys wong Sep 5 '16 at 11:26
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    $\begingroup$ Indeed, so what is $v(x,y)$ ? $\endgroup$ – H. H. Rugh Sep 5 '16 at 11:50
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    $\begingroup$ Is that just zero? $\endgroup$ – ys wong Sep 5 '16 at 12:01
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    $\begingroup$ Precisely! So now you go on calculating derivatives of $u$ and compare... $\endgroup$ – H. H. Rugh Sep 5 '16 at 12:13

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