AMC 10 · 2005 · #22

Grade 12 algebra

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

complex-numberscomplex-polar-formroots-of-unityrecursive-sequenceinduction pattern-recognitionconvert-to-algebra ↑ Prerequisites: complex-numberscomplex-polar-formrecursive-sequence

📏 Long solution 💡 3 insights

Problem

A sequence of complex numbers repeatedly conjugates, divides and multiplies by the imaginary unit, starting on the unit circle. After 2005 steps it lands on one. Count the possible starting values.

Pick an answer.

(A)

(B)

(C)

(D)

2005

(E)

$2^{2005}$

How to solve

Strategy Look for a Pattern

Iterating a rule 2005 times is impossible by hand, so the whole problem is finding a closed form. The conjugate is the obstacle, and tool #15 removes it: on the unit circle z = 1/z, which turns the rule into plain squaring. With squaring in hand, tool #5 makes three terms enough to see the shape of z_n and to prove it by induction. Finally tool #4 names the huge exponent N = 2²⁰⁰⁵, so the whole 2005-step chain becomes a single equation whose solutions are easy to count.

1STEP 1

Every term stays on the unit circle

Absolute values show every term stays on the unit circle.

|z_n+1| = (|i|·|z_n|)/|z_n| = (1·|z_n|)/|z_n| = 1

2STEP 2

Trade the conjugate for a reciprocal

There a conjugate is just a reciprocal, collapsing the rule to a squaring.

z_n+1 = (i z_n)/z_n = (i z_n)/(1/z_n) = i z_n²

3STEP 3

Iterate and watch the constant settle

Iterating doubles the exponent and the constant settles.

z₁ = i z₀², z₂ = i(i z₀²)² = -i z₀⁴, z₃ = i(-i z₀⁴)² = -i z₀⁸

4STEP 4

Prove the closed form

Induction proves a clean closed form.

z_n = -i z₀^{2ⁿ} for all n ≥ 2

5STEP 5

Feed in n = 2005

The whole chain becomes one power equation.

-i z₀^{2²⁰⁰⁵} = 1 ⟹ z₀^{2²⁰⁰⁵} = 1/-i = i

6STEP 6

Count the roots

Counting its roots gives 2²⁰⁰⁵, choice (E).

N = 2²⁰⁰⁵, θ = ((4k+1)π)/2N, k = 0, 1, …, N-1

Answer

2²⁰⁰⁵

The closed form survives a direct spot check: z₂ = i z₁² = i(i z₀²)² = -i z₀⁴, exactly -i z₀^{2²}. The count is also structurally forced — every step squares, so unwinding 2005 steps gives an equation of degree 2²⁰⁰⁵, and none of the smaller options 1, 2, 4, 2005 could match a degree that large. Nothing is lost to the hypothesis |z₀| = 1 either, since every root of z₀^N = i automatically has modulus 1, and nothing extra is gained, since the recursion keeps every term on the unit circle.

💡Key takeaway

Dividing a unit-length complex number by its own conjugate is just squaring it, and squaring 2005 times doubles the exponent every single time.

Every term stays on the unit circle
Trade the conjugate for a reciprocal
Iterate and watch the constant settle
Prove the closed form
Feed in n = 2005
Count the roots