AMC 10 · 2019 · #21
Grade 11 algebraPick an answer.
Tool #9 (Easier Related Problem): nobody computes z¹⁴⁴ directly. Compute z² instead — it is a one-line expansion, and every other power is built from it. Tool #5 (Look for a Pattern): those small powers show that z⁸ = 1, so the powers of z run in a cycle of length 8 and only the exponent's remainder mod 8 matters. The exponents are the perfect squares, and perfect squares mod 8 follow their own short repeating block. Tool #15 (Organize in More Ways): instead of reading the sum left to right, sort the twelve terms into groups by remainder — most of them cancel. Tool #7 (Subproblems): handle the two factors separately, then multiply the two short results at the end.
Read z as a + bi
Read z as real and imaginary parts.
A number of length 1 never grows or shrinks when raised to a power, so its powers can only spin around — which is exactly what makes a cycle possible.
11.N-CN.A.1Organize Information In More WaysSquare it to kill the root
Squaring kills the root.
The √(2) in the denominator is there precisely so that squaring wipes it out and leaves plain i.
11.N-CN.A.2Solve An Easier Related ProblemPowers repeat every eight
The powers repeat every eight.
Once a power comes back to 1, everything after it is a rerun of what already happened.
Once a power comes back to one, everything after it is a rerun of what already happened.
▸ Why?
A complex number of length one only spins as it is raised to powers, so it can return to where it began.
▸ Why?
Once a state repeats, everything that followed it the first time follows it again in the same order.
Squares mod 8: 1, 4, 1, 0
The squares hit only four remainders.
Bumping k by 4 changes k² by a multiple of 8, so the remainders loop every four terms.
6.NS.B.2Look For A PatternGroup the first sum
The first sum collapses to one term.
Sorting the twelve terms by remainder turns a long sum into three piles, and two of the piles cancel.
9.A-SSE.A.2Organize Information In More WaysThe second sum mirrors it
The second sum mirrors it.
1/z is just z flipped across the real axis, so it runs through the same eight-step cycle backwards.
11.N-CN.A.3Identify SubproblemsMultiply and cancel
Multiplying cancels down to 36.
The two sums are mirror images, so whatever direction one points, the other undoes it and only the size 6 × 6 is left.
9.A-SSE.A.2Identify SubproblemsYou never need De Moivre here — just square z to get i, notice z⁸ = 1 so powers repeat every 8, list the remainders of 1, 4, 9, …, 144 mod 8 as the repeating block 1, 4, 1, 0, let the +1 and -1 terms cancel, and the two giant sums shrink to 6z and 6/z, whose product is 36.
- Read z as a + bi
- Square it to kill the root
- Powers repeat every eight
- Squares mod 8: 1, 4, 1, 0
- Group the first sum
- The second sum mirrors it
- Multiply and cancel