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Using Polar Coordinates to Solve a Complex Number Problem

Date: 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)

   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 

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 
Associated Topics:
High School Imaginary/Complex Numbers

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