Analytical Functions

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Homework Statement



Key in writing if possible f (z) with Onley z this mean We can get rid z bar be variable in terms of analytical
Is there a theory or conclude that Ithbt


Where was this idea
Or is the only conclusion
 

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  • #2
malawi_glenn
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What?!?!
 
  • #3
Dick
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I'll back malawi_glenn on that. It's really incoherent. But in x+iy, x and y are two independent variables. In the same way, z and zbar are two independent variables. But you are going to have ask a much clearer question before anyone can even figure out what you are talking about.
 
  • #4
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I'm sorry the question is
Without the use of Kochi - Riemann's equation
Analytical Function:
Example:
[url=http://www.l22l.com][PLAIN]http://www.l22l.com/l22l-up-3/9cfee20d72.bmp[/url][/PLAIN]
 
  • #5
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[url=http://www.x66x.com][PLAIN]http://www.x66x.com/download/10584854dac91b88c.bmp[/url][/PLAIN]
 
  • #7
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[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/5f965970a0.jpg[/url][/PLAIN]
 
  • #8
Dick
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[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/a8bbd62a7c.jpg[/url]
http://www.up07.com/up7/uploads/f733271dae.jpg
Yes. It's general. d/d(zbar)=0 is the same thing as saying i*d/dx=d/dy using the chain rule for partial derivatives. If you apply that to f=u(x,y)+i*v(x,y) you get the Cauchy-Riemann equations.
 
  • #9
Dick
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[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/5f965970a0.jpg[/url][/PLAIN]
Basically ok. (n*i)^(1/2)=sqrt(n/2)+i*sqrt(n/2). Recheck the sqrt(5i). But remember to be careful how you define 'sqrt' or remember that every nonzero number has two different square roots.
 
  • #10
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Yes. It's general. d/d(zbar)=0 is the same thing as saying i*d/dx=d/dy using the chain rule for partial derivatives. If you apply that to f=u(x,y)+i*v(x,y) you get the Cauchy-Riemann equations.
Good Answer
Thanks
 
  • #11
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Basically ok. (n*i)^(1/2)=sqrt(n/2)+i*sqrt(n/2). Recheck the sqrt(5i). But remember to be careful how you define 'sqrt' or remember that every nonzero number has two different square roots.


(n*i)^(1/2)=sqrt(n/2)+i*sqrt(n/2).

Excellent
 

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