### A Problem in Complex Numbers Sets 3 Aside

Kunihiko Chikaya has posted an elegant problem on a facebook page which is a simple exercise in complex numbers that also submits - due to an insightful observation - to a solution with a very little computational effort.

The problem is

Let be complex numbers that satisfy

Find the value of the expression

Denoting leads to a quadratic equation with roots It is now hard to miss the fact that, according to De Moivre's formula, From here, since

However, one solution - submitted by Rozeta Atanasova - obtained the same result without so much as even writing the quadratic equation, let alone solving it. Rozeta Atanasova just observed that immediately implying

This led to the inquiry of answering Kunihiko Chikaya question without solving any equation. Here's one such method. I'd be curious to learn of any other. Algebraically, it takes more steps than that of Rozeta Atanasova, but is still based on a simple insight into the implied properties of the original condition. Let's rewrite it as Squaring gives or Next,

so that The same way, Thus and, by induction, It follows that implying as the answer to Kunihiko Chikaya's question.

With other exponents the situation is more involved. For example,

etc.

The same behavior is observed for the factors of and :

if n is divisible by 3, and otherwise. And

if n is divisible by 3, and otherwise. The latter confirms the answer to the original problem.

In general,

Which clearly makes to stand out.

A shorter proof without solving any equation.

February 16th, 2016 at 9:16 amAs above, x + 1/x = -1 implies x^2 + 1/x^2 = (x + 1/x)^2 - 2 = -1. Now

x + 1/x = x^2 + 1/x^2 is equivalent to x(x-1) = (x-1)/x^2, that is, x^3 = 1.

Thus x^n+ 1/x^n is equal to 1 + 1 or to x + 1/x = x^2 + 1/x^2 = -1 according as n is a multiple of 3 or not.

Yes, I agree. The blog has built-in software that instructed me to make a post of at least 400 words, claiming that, otherwise, google will pass it unnoticed 🙂

February 16th, 2016 at 10:24 am