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Check if the complex function is differentiable

  • Thread starter Fabio010
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  • #1
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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)?
 

Answers and Replies

  • #2
529
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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
FactChecker
Science Advisor
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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
529
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Division by zero is not allowed in complex analysis, so your final equations are not defined at x=y=0. They are not equal.
That is true, but this function is differentiable at ##z=0##. If you evaluate the two limits along the real and imaginary axes (with ##h\in\mathbb{R}##)

[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+0i+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 Cauchy-Riemann 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 Cauchy-Riemann equations are not applicable?
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
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That is true, but this function is differentiable at ##z=0##. If you evaluate the two limits along the real and imaginary axes (with ##h\in\mathbb{R}##)

[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+0i+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 Cauchy-Riemann 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 Cauchy-Riemann equations are not applicable?
I would use the definition of the derivative as a difference quotient to show it's differentiable at z=0.
 
  • #6
529
28
Yes, but ##w(0)=0##, so I left it out.
 
  • #7
Dick
Science Advisor
Homework Helper
26,258
618
Yes, but ##w(0)=0##, so I left it out.
Sure, I'm just saying there is no need work along any particular axes. The complex derivative f'(0) is defined by the limit |h|->0 of (f(0+h)-f(0))/h where h is complex. You can also conclude that that is zero.
 

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