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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)
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
```
Associated Topics:
High School Imaginary/Complex Numbers

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