2002 AMC 12A Problems/Problem 24

Problem

Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$.

$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution 1

Let $s=\sqrt{a^2+b^2}$ be the magnitude of $a+bi$. Then the magnitude of $(a+bi)^{2002}$ is $s^{2002}$, while the magnitude of $a-bi$ is $s$. We get that $s^{2002}=s$, hence either $s=0$ or $s=1$.

For $s=0$ we get a single solution $(a,b)=(0,0)$.

Let's now assume that $s=1$. Multiply both sides by $a+bi$. The left hand side becomes $(a+bi)^{2003}$, the right hand side becomes $(a-bi)(a+bi)=a^2 + b^2 = 1$. Hence the solutions for this case are precisely all the $2003$rd complex roots of unity, and there are $2003$ of those.

The total number of solutions is therefore $1+2003 = \boxed{2004}$.

Solution 2

As in the other solution, split the problem into when $s=0$ and when $s=1$. When $s=1$ and $a+bi=\cos\theta+i\sin\theta$,

$(a+bi)^{2002}= \cos(2002\theta)+i\sin(2002\theta)$ $=a-bi= \cos\theta-i\sin\theta= \cos(-\theta)+i\sin(-\theta)$

so we must have $2002\theta=-\theta+2\pi k$ and hence $\theta=\frac{2\pi k}{2003}$. Since $\theta$ is restricted to $[0,2\pi)$, $k$ can range from $0$ to $2002$ inclusive, which is $2002-0+1=2003$ values. Thus the total is $1+2003 = \boxed{\textbf{(E)}\  2004}$.

Solution 3

Notice that r=0 or r=1 for this to be true. We know this because we are taking magnitude to the 2003rd power, and if the magnitude of a+bi is larger than 1, it will increase and if it is smaller than 1 it will decrease. However, the magnitude on the RHS is still r, so this is not possible. Again, only r=0 and r=1 satisfy.

Now if r=0, we must have (0,0) for (a,b). No exceptions.

However if r=1, we then have:

$cos(2002 \theta) = cos(-\theta)$. This has solution of $\theta = 0$. This would represent the number 1+0i, with conjugate 1-0i. This works because the magnitude is the same and the angle is nothing anyways. We multiply angle by 2002 due to DeMoivre's Theorem and also we do $-\theta$ because it is a reflection, angles therefore is negative.

We then write:

$cos(2002 \theta) = cos(360-\theta)$ which has solution of $\theta = \frac{360}{2003}$.

We can also write:

$cos(2002 \theta) = cos(720-\theta)$ which has solution $\theta = \frac{720}{2003}$.

We notice that it is simply headed upwards and the answer is of the form $\frac{720}{2003} n$, where n is some integer from 0 to infinity inclusive.

Well wait, it repeats itself n=2003, that is 360 which is also 0! Hence we only have n=0 to 2002 as original solutions, or 2003 solutions.

1+2003 = $\boxed{2004}$.

Solution by Blackhawk 9-10-17

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS