Problem 2 Complex Numbers

Problem 2: Complex Numbers

AIME 2008I #8

1 1 1 1 π + tan−1 + tan−1 + tan−1 = . 3 4 5 n 4 √ Definition: Given a complex number z = a + bi, let r = |z| = a2 + b2 . Then θ = arg(z), the argument of z, is the unique angle mod2π such that z = reiθ . Problem: Find the positive integer n such that tan−1

Lemma: arg(zw) ≡ arg(z) + arg(w) mod 2π. Proof: This is actually tangent summation in disguise. It’s just much easier to work with complex numbers than with tangent summation, which is why my solution is short. Lemma: With the appropriate restraints on quadrants, arg(z) = arctan ab . Solution: We can therefore rewrite the problem as arg(3 + i) + arg(4 + i) + arg(5 + i) + arg(n + i) = arg(m + mi) for some real number m and use the first lemma. (3 + i)(4 + i)(5 + i)(n + i) = m + mi =⇒ (11 + 7i)(5 + i)(n + i) = m + mi =⇒ (48 + 46i)(n + i) = m + mi =⇒ 48n − 46 = 46n + 48 =⇒ n = 47

Solution was written by Qiaochu Yuan and compiled from Art of Problem Solving Forums.