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$Old$ April 30th, 2008, 10:40 AM

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$Default$ complex conjurgates calculus of variations

f(z)=√(|xy| )
|z|=√(x^2+y^2 ) in general
Now will that mean
f(z)=√((xy)^(1⁄2) )=xy ??

I will appreciate your help,

Thanks,

Jas

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$Old$ May 1st, 2008, 12:23 AM

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Hello,

Quote:

Originally Posted by JD007 $View Post$

|z|=√(x^2+y^2 ) in general

This is true only if z=x+iy (or z=y+ix)

Quote:

Now will that mean
f(z)=√((xy)^(1⁄2) )=xy ??

If x and y are complex numbers :
Suppose that x=a+bi, and y=c+di.

xy=(a+bi)(c+di)=ac-bd+adi+cbi=(ac-bd)+i(ad+cb)

Hence $|xy|=\sqrt{(ac-bd)^2+(ad+bc)^2}=\sqrt{(ac)^2+(bd)^2+(ad)^2+(bc)^2}$

If x and y are real numbers :
It only means that |xy| is the absolute value of xy. I think it's more appropriate here, but I don't know the context

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