AMC 10 · 2023 · #12

Grade 11 algebra
complex-numberssystems-of-equationsquadratic-equationsabsolute-value convert-to-algebraidentify-subproblemslogical-deduction ↑ Prerequisites: complex-numberssystems-of-equations
📏 Medium solution 💡 3 insights
Problem
A new operation is invented for complex numbers. Instead of mixing the parts as ordinary multiplication does, it multiplies the two real parts together and the two imaginary coefficients together, gluing the second product back on as the imaginary part. Some complex number satisfies that this operation on itself equals its ordinary square plus forty. Find its modulus.

Pick an answer.

(A)
2
(B)
5
(C)
$\sqrt{5}$
(D)
$\sqrt{10}$
(E)
$5\sqrt{2}$
How to solve
Strategy Introduce a Variable

The operation ⊗ is defined piece by piece, in terms of the real part and the imaginary coefficient of each input. So it says nothing at all while z is still a single letter. Tool #4 (Introduce a Variable) is therefore the entire opening move: write z=a+bi with a and b real, and both sides of the equation turn into something you can actually compute. Tool #15 (Organize Information in More Ways) does the next lift: one equation between complex numbers is secretly two equations between real numbers, and you only see them once each side is sorted into a real part and an imaginary part. Tool #7 (Identify Subproblems) keeps the algebra small, because those two equations do not have to be solved together — the real-part equation hands over b² all by itself, and only then does the imaginary-part equation tie a to b. Tool #3 (Eliminate Possibilities) finishes the second equation: factoring it produces two branches, b=0 and b=2a, and one of them has to be ruled out before you can move on.

1STEP 1

Write it with parts showing

The new operation does not mix.

z ⊗ z = (a)(a)+(b)(b)i = a²+b²i
2STEP 2

Square it the ordinary way

Ordinary squaring mixes the parts.

z² = (a+bi)² = (a²-b²)+2ab i ⟹ z²+40 = (a²-b²+40)+2ab i
3STEP 3

Split one equation into two

It becomes two real equations.

a²+b²i = (a²-b²+40)+2ab i ⟹ a² = a²-b²+40 ; b² = 2ab
4STEP 4

What the real part gives

The real part hands over one square.

a² = a²-b²+40 ⟹ 0 = -b²+40 ⟹ b²=40
5STEP 5

What the imaginary part gives

The imaginary part pins the rest.

b²-2ab=0 ⟹ b(b-2a)=0; b²=40 ≠ 0 ⟹ b=2a ⟹ b²=4a²=40 ⟹ a²=10
6STEP 6

Assemble the modulus

Assembling gives five root two.

|z| = √(a²+b²) = √(10+40) = √(50) = √(25 · 2) = 5√(2) (E)
Answer
5√(2)
Build an actual z and test it against the original wording. From a²=10, b²=40 and b=2a, take a=√(10) and b=2√(10), so z = √(10)+2√(10) i. Left side, by the definition: z ⊗ z = a²+b²i = 10+40i. Right side, the ordinary way: z² = (10-40)+2(√(10))(2√(10))i = -30+40i, so z²+40 = 10+40i. The two sides agree, so this z genuinely satisfies the condition. Its modulus is √(10+40) = √(50) = 5√(2) ≈ 7.07. The opposite sign choice a=-√(10), b=-2√(10) also works and has the same modulus, so the answer is not ambiguous. A size check settles it even faster: b²=40 on its own forces |z| = √(a²+b²) ≥ √(40) ≈ 6.32, which is already too big for (A) 2, (B) 5, (C) √(5) and (D) √(10), leaving only (E). Choice (D) √(10) is the trap for anyone who computes a²=10 and stops one line early.
💡Key takeaway

When a problem invents its own operation, write the number as a+bi and follow the definition literally — then match real parts with real parts and imaginary with imaginary, and the single equation splits into two easy ones.

  • Write z with its parts showing
  • Square z the ordinary way
  • Split one equation into two
  • The real part hands over b squared
  • The imaginary part pins down a squared
  • Assemble the modulus