Olympiad Maths Prep

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Problem 741

AIME late
Algebra Difficulty 5.3 Prove it

182. Prove that when two conjugate complex numbers are squared and cubed, the results are again conjugate complex numbers.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let a+bia+b i and a−bia-b i be conjugate complex numbers. Then

(a+bi)2=(a2−b2)+2abi(a−bi)2=(a2−b2)−2abi \begin{aligned} & (a+b i)^{2}=\left(a^{2}-b^{2}\right)+2 a b i \\ & (a-b i)^{2}=\left(a^{2}-b^{2}\right)-2 a b i \end{aligned}

The complex numbers (1) and (2) are conjugates. Similarly, the second statement can be proven.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.