Using Polar Coordinates to Solve a Complex Number ProblemDate: 02/03/2005 at 13:41:50 From: Julie Subject: complex numbers I am a high school math teacher and have given my students a problem from an old AMC contest and I am not sure how to solve it or explain it. Here it is: Find the number of ordered pairs of real numbers (a,b) such that (a + bi) ^ 2002 = a - bi. I know a + bi and a - bi are conjugates and I know how to expand a binomial. I think it is related to powers of i like i^4 = 1 so i^6 = i^2 = -1. The correct answer is 2004. Help!! - Julie Date: 02/03/2005 at 14:19:56 From: Doctor Douglas Subject: Re: complex numbers Hi Julie. This problem is much easier using polar coordinates. If you haven't introduced this yet to your students, it will be excellent motivation. Let's notate a complex number z by z = a + bi = R exp(i Q), where R = sqrt(a^2 + b^2) is the distance of z from the origin 0 + 0i, and Q is the argument: Q = arctan(b/a), where 0 <= Q < 2*pi. In polar coordinates, the given statement is [R exp(i*Q)]^2002 = R exp(-i*Q). Clearly R=0 is a solution: z = 0. Multiplying through by R*exp(i*Q) gives R^2003 exp(i*2003*Q) = R^2, and for nonzero R, we have R^2001 exp(i*2003*Q) = 1. It's clear that R=1 (do you see why?), and we are left with exp(2003*i*Q) = 1. Now, how many values of Q lead to this equation being true? In the allowed range of Q, we must have 2003*Q = 2*pi*N Q = 2*pi*N/2003 so that Q = 0, 2*pi/2003, 4*pi/2003, ... , 2*pi*2002/2003 are all valid values that satisfy the given condition. There are 2003 values in this list. Hence the number of solutions is 2004: one is at the origin z=0 and the others are the 2003 "kth roots of unity", where k=2003. The symmetry of the problem makes it particularly convenient to work things out in polar coordinates--it would be a mess trying to write out things in terms of real and imaginary parts. You may wish to search our archives for other answers using the keyphrase "roots of unity". I hope that this helps! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 02/03/2005 at 14:51:53 From: Julie Subject: complex numbers Thank you for responding and for your excellent solution. I should have thought of that myself! - Julie |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/