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        riddles >> putnam exam (pure math) >> Complex Principal Square Roots (8/26/2002)
        
(Message started by: william wu on Aug 26^th, 2002, 10:10pm)

Title: Complex Principal Square Roots (8/26/2002)
Post by william wu on Aug 26^th, 2002, 10:10pm

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 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: Complex Square Roots
Post by william wu on Aug 27^th, 2002, 7:04pm

To start, here's how to find the square root of a complex number a + bi:

Let w = c + di be the square root of a + bi. Then:

c + di = sqrt(a + bi)

Squaring both sides:

(c + di)^2 = a + bi
c^2 + 2cdi - d^2 = a + bi

This gives us two equations in terms of c and d:

c^2 - d^2 = a
2cd = b

Solve for c and d to get the square root, c + di.

Title: Re: Complex Principal Square Roots
Post by Yournamehere on Aug 29^th, 2002, 2:16pm

on 08/26/02 at 22:10:35, william wu wrote:

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 same sign to Im(sqrt(z)) as to Im(z). (Yes, zero has a sign.)

What exactly does "Real Part is same sign to Im(sqrt(z)) as to Im(z)" mean? Does this mean signum(Im(sqrt(z))) = signum(Im(z)), or does it mean signum(Re(sqrt(z)))*signum(Im(sqrt(z))) = signum(Re(sqrt(z)))*signum(Im(z)), or does it mean something else?

Title: Re: Complex Principal Square Roots
Post by william wu on Sep 3^rd, 2002, 6:51pm

Sorry for the wording; I was aware that it was unclear but that's copied verbatim from the blackboard.

I think it means that sign(Re(sqrt(z))) = sign(Im(sqrt(z))) = sign(Im(z)).

Honestly I'm not sure about this problem.

Title: Re: Complex Principal Square Roots
Post by Yournamehere on Sep 4^th, 2002, 10:25am

That definition can't be right. Neither square root of -i, for instance, satisfies signum(Re(sqrt(-i))) = signum(Im(sqrt(-i))).

The straightforward approach is simply to enumerate cases. Both f and g (and F and G) are both zeros of some polynomial z^2+k. From the Fundamental Theorem of Algebra, there are only two complex zeros, and it's easy to show that if the zeros are a and b, then, in polar notation, |a| = |b| and Arg(a) = Arg(b)+\pi. So you only need to determine whether f(z) and g(z) both lie in some half of the Cartesian plane or not [1]. If they do lie in the same half-plane, then they must be equal; if not, they must be different. This is much easier than computing f(z) and g(z) explicitly.

Exactly what the cases are to split on depends on the definition of Principal Square Root. The usual definition I understood is -\pi/2 < Arg(Sqrt(z)) <= \pi/2, but I cannot make this correspond at all with the definition given in the problem. I'll post a solution based on this definition later.

[1] Technically you can't use the usual definition of a half-plane (open or closed); the plane must include half of its boundary points. Sort of "half-open, half-closed". -\pi/2 < Arg(z) <= \pi/2 is an example of such.

Title: Re: Complex Principal Square Roots
Post by Yournamehere on Sep 4^th, 2002, 10:07pm

Looking back at what I wrote this morning, I think I could have been clearer and said "g^2=f^2, so either g=f or g=-f". Anyways, let's assume Principal Square Root is defined by -\pi/2 < Arg Sqrt(z) <= \pi/2. Recall that for any two complex numbers s, t, Arg (st) = Arg s + Arg t, and also Arg Sqrt(z) = (Arg z)/2. Now consider the following cases:

Case 1: Im z > 0. Then Im (1+z) > 0, so 0 < Arg (1+z) < \pi, and 0 < Arg Sqrt(1+z) < \pi/2. Also Im (1-z) < 0, so -\pi < Arg (1-z) < 0, and -\pi/2 < Arg Sqrt(1-z) < 0. Thus -\pi/2 < Arg (Sqrt(1-z) Sqrt(1+z)) < pi/2. Since -\pi/2 < Arg f <= \pi/2, g cannot equal -f, so g=f.

Case 2: Im z < 0. Similar reasoning shows g=f as well.

Case 3: Im z = 0. Easy to show g=f, since now 1-z and 1+z are guaranteed to be real.

So g=f for all z.

Part (b) is a bit trickier. Note that the above analysis does not give a clear answer, for although 0 < Arg Sqrt(z+1) < \pi/2, we have Im (z-1) > 0, so 0 < Arg Sqrt(z-1) < \pi/2, and hence 0 < Arg (Sqrt(z-1) Sqrt(z+1)) < \pi. So either 0 < Arg G <= \pi/2 (and thus we must have G=F), or \pi/2 < Arg G < \pi, and G must be -F. So we need to do something to further distinguish these two outcomes. I'll only give a hint to the rest, in that it might be difficult to put further bounds on Arg G, and that it might be easier to find other ways to determine if G=F.

I'm not sure how friendly a message board like this is to discussing math problems; formatting and notation can be difficult to convey.

Title: Re: Complex Principal Square Roots
Post by Yournamehere on Sep 5^th, 2002, 10:05am

on 09/04/02 at 22:07:39, Yournamehere wrote:

Note that the above analysis does not give a clear answer, for although 0 < Arg Sqrt(z+1) < \pi/2, we have Im (z-1) > 0, so 0 < Arg Sqrt(z-1) < \pi/2, and hence 0 < Arg (Sqrt(z-1) Sqrt(z+1)) < \pi.

Oops. I should have noted that this refers to "Case 1" only. Case 2 runs into a similar problem, though.