AMC 10 · 2023 · #14

Grade 12 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

complex-numberscomplex-polar-formroots-of-unityabsolute-value caseworkidentify-subproblems ↑ Prerequisites: complex-numbersexponents

📏 Medium solution 💡 3 insights

Problem

Count the complex numbers whose fifth power equals their own conjugate.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Identify Subproblems

The conjugate blocks every polynomial technique, so Tool #9 (Solve an Easier Related Problem) makes the first move: compare only the sizes of the two sides. That collapses the whole complex plane into two possible moduli, |z|=0 and |z|=1, which is exactly the split Tool #7 (Identify Subproblems) is for — two disjoint cases, counted separately and then added. Tool #4 (Introduce a Variable) names the modulus r = |z| and, in the second case, writes z in polar form so the conjugate becomes 1/z. Tool #2 (Make a Systematic List) finishes by writing out every root explicitly, which is the only way to be sure nothing is double counted and nothing is dropped.

1STEP 1

Compare the sizes

Compare the moduli.

|z⁵| = |z| ⟹ |z|⁵ = |z| ⟹ r⁵ = r, r = |z| ≥ 0

2STEP 2

Solve for the modulus

The modulus is zero or one.

r⁵ - r = r (r-1)(r+1)(r²+1) = 0, r ≥ 0 ⟹ r = 0 or r = 1

3STEP 3

Check the origin

The origin counts too.

z = 0: z⁵ = 0, z = 0 ⟹ 1 solution

4STEP 4

Move to the unit circle

On the unit circle the conjugate is the reciprocal.

|z| = 1 ⟹ z = 1/z; z⁵ = 1/z ⟺ z⁶ = 1

5STEP 5

Count the unit-circle roots

Six solutions appear.

z = exp ((2π i k)/6), k = 0,…,5: 1, 1/2+√(3)/2i, -1/2+√(3)/2i, -1, -1/2-√(3)/2i, 1/2-√(3)/2i

6STEP 6

Add them up

Together that is 7.

1 + 6 = 7 ⟹ (E) 7

Answer

Cross-check by a route that never uses the case split. Conjugating z⁵ = z gives z^ 5 = z, and substituting z = z⁵ into that yields (z⁵)⁵ = z, so z²⁵ = z and z (z²⁴-1) = 0. Every solution therefore sits among only 25 candidates: z = 0 and the twenty-four 24th roots of unity. Testing z = exp ((2π i k)/24) in the original equation needs exp(5iθ) = exp(-iθ), i.e. 6θ a whole number of turns, which happens exactly when k is a multiple of 4 — that is 6 of the 24. Adding z = 0 gives 7 again, from a completely independent direction. The list also matches the geometry: a regular hexagon on the unit circle plus its center. Note that choice (D) 6 is precisely the trap for a solver who multiplies by z at the very start and silently loses z = 0.

💡Key takeaway

Compare sizes first: |z|⁵ = |z| forces |z| = 0 or |z| = 1. The origin is one solution on its own, and on the unit circle z = 1/z turns the equation into z⁶ = 1 with six roots, so the total is 1 + 6 = (E) 7. The point everyone drops is z = 0.