AMC 10 · 2021 · #18

Grade 11 algebra
complex-numberscompleting-the-squareabsolute-valuesymmetric-polynomials convert-to-algebraextreme-principleidentify-subproblems ↑ Prerequisites: complex-numberscompleting-the-square
📏 Long solution 💡 4 insights
Problem
A complex number satisfies one equation built from squared moduli. Find the value of that number plus six divided by it.

Pick an answer.

(A)
-2
(B)
-1
(C)
$\frac12$
(D)
1
(E)
4
How to solve
Strategy Introduce a Variable

Absolute-value bars block algebra, so Tool #15 (Organize Information in More Ways) rewrites each |w|² as the product ww — the same quantity, now expandable. Once that is done, every expansion produces only two combinations, z + z and zz, because the equation cannot distinguish z from its conjugate. Tool #4 (Introduce a Variable) names those two combinations s and p; both are real, so Tool #13 (Convert to Algebra) turns a statement about a complex number into one ordinary real equation in s and p. Completing the square collapses that equation to a sum of two squares equal to zero, and Tool #14 (Extreme Principle) finishes it: a square is never smaller than zero, so a zero total forces each square to be zero and pins s and p exactly.

1STEP 1

Trade moduli for conjugate products

Rewrite each as a conjugate product.

|z|² = zz, |z+2|² = (z+2)(z+2), |z²+1|² = (z²+1)(z²+1)
2STEP 2

Name the two real quantities

Both the sum and the product are real.

s = z + z ∈ R, p = zz = |z|² ∈ R, z² + z² = s² - 2p
3STEP 3

Expand each term

Write every term with those two.

|z+2|² = p + 2s + 4, |z²+1|² = p² + (s² - 2p) + 1
4STEP 4

Assemble one real equation

Every imaginary part disappears.

12p = p² + s² + 4s + 40 ⟹ p² - 12p + s² + 4s + 40 = 0
5STEP 5

Complete the square twice

Two perfect squares appear.

(p-6)² - 36 + (s+2)² - 4 + 40 = 0 ⟹ (p-6)² + (s+2)² = 0
6STEP 6

A zero sum forces both

The zero sum forces each to vanish.

(p-6)² ≥ 0, (s+2)² ≥ 0, sum = 0 ⟹ p = |z|² = 6, s = z + z = -2
7STEP 7

Read off the requested value

The value is negative two.

z + 6/z = z + zz/z = z + z = s = -2 ⟹ (A)
Answer
-2
First check that such a z exists. Writing z = a + bi with a and b real, z + z = 2a = -2 gives a = -1, and |z|² = a² + b² = 6 gives b² = 5. So z = -1 ± i√(5), a genuine complex number, and the two conditions are consistent rather than contradictory. Now test z = -1 + i√(5) in the original equation: |z|² = 1 + 5 = 6, so the left side is 12 · 6 = 72; |z+2|² = |1 + i√(5)|² = 1 + 5 = 6, contributing 2 · 6 = 12; and z² = -4 - 2i√(5), so |z²+1|² = |-3 - 2i√(5)|² = 9 + 20 = 29. The right side is 12 + 29 + 31 = 72, matching. Finally 6/(-1 + i√(5)) = (6(-1 - i√(5)))/6 = -1 - i√(5), so z + 6/z = -2, which is choice (A). The conjugate root -1 - i√(5) returns the same value, as it must: the answer choices are all real, so both roots have to agree.
💡Key takeaway

Replace each |w|² by ww, name the two real quantities z + z and zz, complete the square until a sum of squares equals zero — then 6/z is simply z.

  • Trade moduli for conjugate products
  • Name the two real quantities
  • Expand each term in s and p
  • Assemble one real equation
  • Complete the square twice
  • A zero sum forces both squares
  • Read off the requested value