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Check if the complex function is differentiable 
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#1
Jan2614, 09:41 AM

P: 83

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)? 


#2
Jan2614, 09:58 AM

P: 460

If you just plug in ##y=0## and ##x=0## you will get an indeterminate form which is meaningless. If you evaluate the limits, I think that you get all expressions equal to ##0##, but double check that.



#3
Jan2614, 07:34 PM

P: 357

Division by zero is not allowed in complex analysis, so your final equations are not defined at x=y=0. They are not equal.



#4
Jan2714, 10:00 AM

P: 460

Check if the complex function is differentiable
[itex] \displaystyle\lim_{h\rightarrow0}\displaystyle\frac{(0+0i+h)\left(0+0i +h)\right}{h}=0 [/itex] [itex] \displaystyle\lim_{h\rightarrow0}\displaystyle\frac{(0+0i+ih)\left(0+0 i+ih)\right}{ih}=0 [/itex] So the function is differentiable at ##0##. I don't remember enough from my complex analysis course (which had a number of students who had not taken real analysis, so it was a bit less rigorous than some courses) to reconcile this. My recollection is that the limits of the CauchyRiemann equations could be evaluated, but a quick look online showed that my recollection was incorrect. Perhaps, since the partial derivatives are undefined at 0 the CauchyRiemann equations are not applicable? 


#5
Jan2714, 10:20 AM

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#6
Jan2714, 06:39 PM

P: 460

Yes, but ##w(0)=0##, so I left it out.



#7
Jan2714, 06:59 PM

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