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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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- Thread starter m_s_a
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- #1

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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

- #2

malawi_glenn

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What?!?!

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Dick

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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]

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[url=http://www.x66x.com][PLAIN]http://www.x66x.com/download/10584854dac91b88c.bmp[/url][/PLAIN]

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http://www.up07.com/up7/uploads/f733271dae.jpg

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[url=http://www.up07.com/up7][PLAIN]http://www.up07.com/up7/uploads/5f965970a0.jpg[/url][/PLAIN]

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Dick

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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.

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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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