AMC 10 · 2017 · #12

Grade 12 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
roots-of-unitycomplex-polar-formtrigonometric-ratios symmetry-argumentsystematic-enumeration ↑ Prerequisites: complex-polar-formroots-of-unity
📏 Medium solution 💡 3 insights
Problem
The equation z to the twelfth equals 64 has twelve complex solutions. Some of them have a real part strictly greater than zero. Add up exactly those solutions and report the total.

Pick an answer.

(A)
2
(B)
4
(C)
$\sqrt{2}+2\sqrt{3}$
(D)
$2\sqrt{2}+\sqrt{6}$
(E)
$(1+\sqrt{3}) + (1+\sqrt{3})i$
How to solve
Strategy Organize Information in More Ways

Written as a + bi, the twelve roots are a pile of square roots that nobody wants to expand and then sort by hand. The same twelve numbers have a second storage format: a size (distance from 0) together with a direction (angle). That format is built for this problem, because raising to the 12th power multiplies sizes and adds angles, so the equation splits cleanly into one easy size condition and one easy angle condition. Once the roots are stored as angles, "positive real part" becomes the visual test "strictly right of the vertical axis", which can be read straight off a picture, and the leftover sum collapses because mirror-image roots cancel their imaginary parts.

1STEP 1

Every root has the same size

All twelve share the same size.

r¹² = 64 = 2⁶ → r = 2⁶/12 = 2¹/2 = √(2)
2STEP 2

Name the twelve roots by angle

The angle is what tells them apart.

z_k = √(2)(cos(30° k) + isin(30° k)), k = 0, 1, 2, …, 11
3STEP 3

Keep the roots pointing right

Only five point rightward.

Re(z_k) = √(2)cos(30° k) ＞ 0 ⇔ 30° k ∈ {0°, 30°, 60°, 300°, 330°}
4STEP 4

Pair mirror roots to kill the i

Mirror pairs make the imaginary parts cancel.

(a + bi) + (a - bi) = 2a, S = √(2) + 2√(2)cos 30° + 2√(2)cos 60°
5STEP 5

Add the real parts

Adding the surviving real parts gives two root two plus root six.

S = √(2)(1 + 2 · √(3)/2 + 2 · 1/2) = √(2)(2 + √(3)) = 2√(2) + √(6)
Answer
2√(2)+√(6)
The two forms of the answer agree numerically: 2√(2) + √(6) ≈ 2.828 + 2.449 = 5.277, and √(2)(2 + √(3)) ≈ 1.414 × 3.732 ≈ 5.277. A size check also fits: five roots are being added and each has real part at most √(2) ≈ 1.414, so the sum must be below 5√(2) ≈ 7.07, and it must exceed √(2) because four more positive real parts are stacked on top of it. That interval kills (A) 2 and (B) 4, and the exact algebra separates the answer from (C) √(2) + 2√(3) ≈ 4.88. Choice (E) falls on structure alone: the survivors are one real root plus two conjugate pairs, so the sum cannot carry an i. One more consistency note — the two excluded roots on the vertical axis are i√(2) and -i√(2), which add to 0, so mistakenly including them would not have changed the total; the mistake that actually costs the problem is letting in a root that points left.
💡Key takeaway

The twelve solutions of z¹² = 64 sit evenly around a circle of radius √(2), so keep only the five that point strictly right, let the mirror pairs cancel their imaginary parts, and add the real parts that are left.

  • Every root has the same size
  • Name the twelve roots by angle
  • Keep the roots pointing right
  • Pair mirror roots to kill the i
  • Add the real parts

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