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Topic: Complex principal square root (Read 2675 times) |
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Ryan H
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The complex Principal Square Root sqrt(z) of a complex number z is the square root (one of two if z != 0) whose Real Part is nonnegative. If z<0 (so Re(sqrt(z)) = 0) then assign the same sign to Im(sqrt(z)) as to Im(z). (Yes, zero has a sign.)
Now let:
f(z) := sqrt(1 - z^2), g(z) := sqrt(1-z) * sqrt(1+z)
F(z) := sqrt(z^2 - 1), G(z) := sqrt(z-1) * sqrt(z+1)
Over what region in the complex plane does f(z) = g(z)?
Over what region in the complex plane does F(z) = G(z)?"
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| « Last Edit: Apr 25th, 2005, 3:36pm by Icarus » |
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towr
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Re: Help plz
« Reply #1 on: Apr 23rd, 2005, 2:47pm » |
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Those functions just describe lines in the complex plane, they overlap in points, not regions.
Or do you mean where they have the same sign?
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Icarus
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Re: Help plz
« Reply #2 on: Apr 24th, 2005, 3:16pm » |
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I'm not quite sure what towr is thinking of. These are functions from C --> C, not lines.
To figure this out, look at the values of various points, and then try to decide where the transitions between equal and not equal occur. In particular, consider each of the points z=1, -1, i, -1, 1+i, 1-i, -1+i, -1-i. They will tell you most of what you need to know.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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VincentLascaux
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If r in R+ and theta in [-Pi, Pi[, sqrt(r e(i*theta)) = sqrt(r) * e^(i theta/2)
(1) <=> f(z) = g(z) <=> sqrt((1-z)*(1+z)) = sqrt(1-z) * sqrt(1+z)
(2) <=> F(z) = G(z) <=> sqrt((z+1)*(z-1)) = sqrt(z+1) * sqrt(z-1)
sqrt(a*b) = sqrt(|a|*|b|) * sqrt(e(i (thetaA+thetaB))
sqrt(a) * sqrt(b) = sqrt(|a|*|b|) * e(i (tethaA + thetaB)/2)
So sqrt(ab) = sqrt(a)*sqrt(b) iif thetaA + thetaB in [-Pi, Pi]
(1) <=> arg(1-z)+arg(1+z) in [-Pi, Pi[
(2) <=> arg(z+1)+arg(z-1) in [-Pi, Pi[
I think I'll stop here... it's getting pretty ugly after that point 
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towr
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Re: Help plz
« Reply #4 on: Apr 24th, 2005, 11:02pm » |
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on Apr 24th, 2005, 3:16pm, Icarus wrote:| I'm not quite sure what towr is thinking of. These are functions from C --> C, not lines. |
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I meant curves.. But anyway..
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VincentLascaux
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Quote:| I meant curves.. But anyway. |
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They are not curves either: if we use the graphical representation of complex, to each point z, they associate another point (f(z) or g(z)).
So they are transformations, not curves
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Deedlit
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Re: Help plz
« Reply #7 on: Apr 25th, 2005, 5:56pm » |
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on Apr 24th, 2005, 5:44pm, VincentLascaux wrote:
(1) <=> arg(1-z)+arg(1+z) in [-Pi, Pi[
(2) <=> arg(z+1)+arg(z-1) in [-Pi, Pi[
I think I'll stop here... it's getting pretty ugly after that point |
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It's not so bad - you're getting there. A little geometric intuition at this point is better than straight algebra crunching; what does it mean for the sum of two arguments to be more than pi? How do you represent the argument geometrically?
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