AMC 10 · 2002 · #24
Grade 12 algebraPick an answer.
Written in a and b, the equation is a pair of monstrous real polynomial equations obtained by expanding a 2002nd power — hopeless to attack directly. Two rearrangements fix that. First, a and b only ever appear in the two combinations a+bi and a-bi, so naming z = a+bi replaces two real unknowns by one complex unknown and turns the problem into z²⁰⁰² = z. Second, the conjugate is the real obstacle: it cannot be expanded, factored, or differentiated away. But a number times its conjugate collapses to |z|², a plain nonnegative real. So multiplying both sides by z converts the equation into the ordinary power equation z²⁰⁰³ = |z|², and that single multiplication is what makes the count possible. From there the size of z and the direction of z separate cleanly: the size equation has just two solutions, and each is a small subproblem. The exponent quietly becomes 2003, not 2002, which is exactly where the answer choices are trying to catch you, so the final count deserves a check on small exponents where every solution can be listed by hand.
Trade two real unknowns for one complex one
Both unknowns appear only as a number and its conjugate, so set z = a+bi and the equation becomes z²⁰⁰² = z̄.
The pair (a,b) and the number a+bi carry identical information, so switching to z neither creates nor destroys solutions.
11.N-CN.A.1Introduce A VariableMultiply by z to erase the conjugate
Multiplying both sides by z turns the conjugate into a real modulus: z²⁰⁰³ = |z|².
A number times its conjugate is just a plain squared length, so one multiplication turns a conjugate equation into a power equation.
Multiplying through by z turns the conjugate into a plain squared length and leaves a pure power equation.
▸ Why?
A complex number is a point, so a number times its conjugate is just the square of its distance from the origin.
▸ Why?
That product is a sum times a difference of the same two parts, which cancels the middle and leaves two squares.
Let size decide first
Matching sizes gives r²(r²⁰⁰¹-1) = 0, so r = 0 or r = 1 and the rest of the plane is gone.
Length obeys its own small equation, and raising a positive length to a huge power returns it to itself only at 0 and 1.
9.A-SSE.A.2Identify SubproblemsCase r = 0
The size-zero case is just the origin, giving the single pair (0,0).
The origin is the one place where a gigantic power still lands back where it started.
11.N-CN.A.1Identify SubproblemsCase r = 1: count the 2003rd roots of unity
On the unit circle the equation reads z²⁰⁰³ = 1, whose roots number 2003.
Raising to a power spins the angle, and landing back on 1 means the spin closed a whole number of full turns.
12.N-CN.B.5Organize Information In More WaysAdd the two cases
The cases share nothing, so the total is 1 + 2003 = 2004, choice (E).
The exponent that actually governs the count is 2003, one more than the one printed, and the origin adds a single extra solution on top.
11.N-CN.C.9Identify SubproblemsWhen an equation mixes a number with its conjugate, multiply through by the number itself: zz collapses into a plain real square, and what is left is an ordinary power equation whose solutions you can count.
- Trade two real unknowns for one complex one
- Multiply by z to erase the conjugate
- Let size decide first
- Case r = 0
- Case r = 1: count the 2003rd roots of unity
- Add the two cases