...Prove Their Worth...

"Problems worthy of attack
prove their worth
by hitting back." - Piet Hein

A kind of running diary and rambling pieces on my struggles with assorted books, classes, and other things, as they happen. You must be pretty bored to be reading this...

Wednesday, June 05, 2002

So. Take a circle on a plane. Make it a complex plane, just for kicks. The unit tangent to the circle is i *z/Length[z]. That's because if our coordinate on the circle is z, then to get a unit vector we can divide by the length of z, and then we turn it by Pi/2 by multiplying by i, and now we've got a unit tangent. Now, we're interested in how the curvature of a given shape changes under an analytic mapping. It's possible to show that if we pick the above-mentioned circle as our 'shape', then the image curvature under an analytic mapping f(z) is given by

k_image = (1 + Re[z*f''/f'])/(z*f')

Now, the question is, without actually cranking out a calculation, what should the image curvature be if our mapping is given by f(z) = z^m? And then check your prediction using the formula.

So. z^m is certainly analytic. Hmm. Well. z^m basically dilates the plane, more or less, right? I mean, it takes each point, and pushes it out z^m. Which suggests that the curvature, which starts out as 1/Length[z] is going to be something like 1/Length[z^m]. Ok.

Doing the calculation, though, leads to a rather different conclusion. The image curvature is instead:

image_curvature = 1/Length[m*z^m] + m*(m-1)/Length[m*z^m]

WTF? The scary thing is that this bizarre result matches intuition if m = 1. Because then it's just 1/Length[z], which is the same as the initial curvature, which damn well better be the case, because m=1 we're just doing a linear mapping, that is, we're not changing jack shit.

I've got a sneaking suspicion I'm misunderstanding how z^m works geometrically, which would be embarassing...


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