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## Main Question or Discussion Point

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

[tex]w=z \left| z\right|[/tex]

[tex]w=\sqrt{x^2+y^2} (x+i y)[/tex]

[tex]u = x\sqrt{x^2+y^2}[/tex]

[tex]v = y\sqrt{x^2+y^2}[/tex]

Using the Cauchy Riemann equations

[tex]\frac{\partial }{\partial x}u=\frac{\partial }{\partial y}v[/tex]

[tex]\frac{\partial }{\partial y}u=-\frac{\partial }{\partial x}v[/tex]

my results:

[tex]\frac{x^2}{\sqrt{x^2+y^2}}=\frac{y^2}{\sqrt{x^2+y^2}}[/tex]

[tex]\frac{x y}{\sqrt{x^2+y^2}}=0[/tex]

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

[tex]w=z \left| z\right|[/tex]

[tex]w=\sqrt{x^2+y^2} (x+i y)[/tex]

[tex]u = x\sqrt{x^2+y^2}[/tex]

[tex]v = y\sqrt{x^2+y^2}[/tex]

Using the Cauchy Riemann equations

[tex]\frac{\partial }{\partial x}u=\frac{\partial }{\partial y}v[/tex]

[tex]\frac{\partial }{\partial y}u=-\frac{\partial }{\partial x}v[/tex]

my results:

[tex]\frac{x^2}{\sqrt{x^2+y^2}}=\frac{y^2}{\sqrt{x^2+y^2}}[/tex]

[tex]\frac{x y}{\sqrt{x^2+y^2}}=0[/tex]

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

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