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Check if the complex function is differentiable

by Fabio010
Tags: check, complex, differentiable, function
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Fabio010
#1
Jan26-14, 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)?
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DrewD
#2
Jan26-14, 09:58 AM
P: 446
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.
FactChecker
#3
Jan26-14, 07:34 PM
P: 311
Division by zero is not allowed in complex analysis, so your final equations are not defined at x=y=0. They are not equal.

DrewD
#4
Jan27-14, 10:00 AM
P: 446
Check if the complex function is differentiable

Quote Quote by FactChecker View Post
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+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 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?
Dick
#5
Jan27-14, 10:20 AM
Sci Advisor
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Quote Quote by DrewD View Post
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+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 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.
DrewD
#6
Jan27-14, 06:39 PM
P: 446
Yes, but ##w(0)=0##, so I left it out.
Dick
#7
Jan27-14, 06:59 PM
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P: 25,228
Quote Quote by DrewD View Post
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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