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The question is to check where the following complex function is differentiable.

$$w=z \left| z\right|$$

$$w=\sqrt{x^2+y^2} (x+i y)$$

$$u = x\sqrt{x^2+y^2}$$

$$v = y\sqrt{x^2+y^2}$$

Using the Cauchy Riemann equations

$$\frac{\partial }{\partial x}u=\frac{\partial }{\partial y}v$$

$$\frac{\partial }{\partial y}u=-\frac{\partial }{\partial x}v$$

my results:

$$\frac{x^2}{\sqrt{x^2+y^2}}=\frac{y^2}{\sqrt{x^2+y^2}}$$

$$\frac{x y}{\sqrt{x^2+y^2}}=-\frac{x y}{\sqrt{x^2+y^2}}$$

solutions says that it's differentiable at (0,0). But doesn't it blow at (0,0)?

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  • $\begingroup$ And I think you made a mistake with $v'_x$. $\endgroup$ – Poppy Jan 26 '14 at 15:43
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Hint: You haven't found the partial derivative $\frac{\partial }{\partial y}u$ at $(0,0)$. Your formulas are true for all points except $(0,0)$.

$\displaystyle u'_y(0,0)=\lim\limits_{h\to0}\frac{u(0,h)-u(0,0)}{h}=\lim\limits_{h\to0}\frac{0-0}{h}=0$

$\displaystyle v'_x(0,0)=\lim\limits_{h\to0}\frac{v(h,0)-v(0,0)}{h}=\lim\limits_{h\to0}\frac{0-0}{h}=0$

And the same for the first C-R equation.

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