Can Complex Analysis Simplify (n*i)^(1/2) Expressions?

[m_s_a]

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

 

[malawi_glenn]

What??

 

[Dick]

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.

 

[m_s_a]

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]

 

[m_s_a]

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[m_s_a]

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[m_s_a]

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[Dick]

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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.

 

[Dick]

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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.

 

[m_s_a]

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

 

[m_s_a]

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