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        riddles >> putnam exam (pure math) >> Complex principal square root
        
(Message started by: Ryan H on Apr 23^rd, 2005, 11:48am)

Title: Complex principal square root
Post by Ryan H on Apr 23^rd, 2005, 11:48am

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)?"

Title: Re: Help plz
Post by towr on Apr 23^rd, 2005, 2:47pm

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?

Title: Re: Help plz
Post by Icarus on Apr 24^th, 2005, 3:16pm

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.

Title: Re: Help plz
Post by VincentLascaux on Apr 24^th, 2005, 5:44pm

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 :)

Title: Re: Help plz
Post by towr on Apr 24^th, 2005, 11:02pm

on 04/24/05 at 15:16:12, Icarus wrote:

I'm not quite sure what towr is thinking of. These are functions from C --> C, not lines.

I meant curves.. But anyway..

Title: Re: Help plz
Post by VincentLascaux on Apr 25^th, 2005, 2:38am

Quote:

I meant curves.. But anyway.

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

Title: Re: Help plz
Post by towr on Apr 25^th, 2005, 4:47am

Yeah sorry.. haven't been getting enough sleep lately ;D

Title: Re: Help plz
Post by Deedlit on Apr 25^th, 2005, 5:56pm

on 04/24/05 at 17:44:31, 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 :)

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?