Tuesday, April 05, 2005

Factor away your troubles!

Saturday night, I was sitting in a dark room listening to some speakers. While they were fairly interesting (one guy started out telling about a drug bust that almost went bad), I found myself getting very bored. I always do when I’m just listening to people talking, so this came as no surprise, but this time I had nothing that I could really do. I had no useable electronics on hand, and it was too dark to use something that didn’t have backlighting. So, I remembered a problem that I had been working on a few days before:

r is a root of x^4-x^3+x^2-x+1=0. Evaluate r^40-r^30+r^20-r^10+1. (The original problem stated that the latter equation was equal to 0, but that was incorrect.)

I had tried a thing or two on paper before, and fallen asleep before making any headway when trying to work it out in my head. While we still had a bit of light from the dusk, I tried out a few more things on paper, but I soon found myself working entirely in my head in the darkness.

Quickly I could eliminate -1, 0, and 1 as possible roots. I did realize though that the product of the four roots is 1, and that their sum is -1. From this, I assumed that all four roots had magnitude 1 in the complex domain. If a root is complex, then the conjugates must cancel. From this, I assumed that the imaginary part of the x^4 term corresponded to the imaginary part of the x term. I let r=e^it where e is Euler’s number, i is the imaginary number, and t is an angle (in radians). If -(e^it)+(e^4it) is real, then pi+t=2*pi-4t. This gives 5t=pi (actually, 5t=pi+2k*pi, where k=0, 1, 3, or 4), and so r=e^i*pi/5, or (back in degrees), r=cos 72+i*sin 72. This means that r^10=1, and the desired solution is 1-1+1-1+1=1.

The cool thing about this is that I correctly factored a quartic function with complex roots in my head, and it took less than an hour, leaving me free to learn about the evils of poker later in the meeting, comforted by the fact that r^40-r^30+r^20-r^10+1=1<>0.

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