# CTK Insights

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15 Feb

### 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 $a,b$ be complex numbers that satisfy $\displaystyle\frac{a}{b}+\frac{b}{a}+1=0.$
Find the value of the expression $\displaystyle\frac{a^{2016}}{b^{2016}}+\frac{b^{2016}}{a^{2016}}+1.$

Denoting $\displaystyle x=\frac{a}{b}$ leads to a quadratic equation $x^2+x+1=0,$ with roots $\displaystyle x_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}=\cos 120^{\circ}\pm i\sin 120^{\circ}.$ It is now hard to miss the fact that, according to De Moivre's formula, $\displaystyle x_{1,2}^3=\cos 360^{\circ}\pm i\sin 360^{\circ}=1.$ From here, since $2016=3\times 672,$ $\displaystyle\frac{a^{2016}}{b^{2016}}+\frac{b^{2016}}{a^{2016}}+1=1+1+1=3.$

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 $(x-1)(x^2+x+1)=x^3-1,$ immediately implying $x^3=1.$

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 $\displaystyle\frac{a}{b}+\frac{b}{a}=-1.$ Squaring gives $\displaystyle\left(\frac{a}{b}\right)^2+2+\left(\frac{b}{a}\right)^2=1,$ or $\displaystyle\frac{a^2}{b^2}+\frac{b^2}{a^2}=-1.$ Next,

$\displaystyle(-1)(-1)=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)-1,$ so that $\displaystyle\frac{a^3}{b^3}+\frac{b^3}{a^3}=2.$ The same way, $\displaystyle\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)=\left(\frac{a^6}{b^6}+\frac{b^6}{a^6}\right)+\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)=\left(\frac{a^6}{b^6}+\frac{b^6}{a^6}\right)+2=4.$ Thus $\displaystyle\frac{a^6}{b^6}+\frac{b^6}{a^6}=2,$ and, by induction, $\displaystyle\frac{a^{3n}}{b^{3n}}+\frac{b^{3n}}{a^{3n}}=2.$ It follows that $\displaystyle\frac{a^{2016}}{b^{2016}}+\frac{b^{2016}}{a^{2016}}=2,$ implying $3$ as the answer to Kunihiko Chikaya's question.

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

$\displaystyle\frac{a^2}{b^2}+\frac{b^2}{a^2}=-1,$ $\displaystyle\frac{a^4}{b^4}+\frac{b^4}{a^4}=-1,$ $\displaystyle\frac{a^6}{b^6}+\frac{b^6}{a^6}=2,$ $\displaystyle\frac{a^8}{b^8}+\frac{b^8}{a^8}=-1,$ $\displaystyle\frac{a^{10}}{b^{10}}+\frac{b^{10}}{a^{10}}=-1,$ $\displaystyle\frac{a^{12}}{b^{12}}+\frac{b^{12}}{a^{12}}=2,$ etc.

The same behavior is observed for the factors of $5$ and $7$:

$\displaystyle\frac{a^{5n}}{b^{5n}}+\frac{b^{5n}}{a^{5n}}=2,$ if n is divisible by 3, and $\displaystyle\frac{a^{5n}}{b^{5n}}+\frac{b^{5n}}{a^{5n}}=-1,$ otherwise. And

$\displaystyle\frac{a^{7n}}{b^{7n}}+\frac{b^{7n}}{a^{7n}}=2,$ if n is divisible by 3, and $\displaystyle\frac{a^{7n}}{b^{7n}}+\frac{b^{7n}}{a^{7n}}=-1,$ otherwise. The latter confirms the answer to the original problem.

In general,

Which clearly makes $3$ to stand out.

#### 2 Responses to “A Problem in Complex Numbers Sets 3 Aside”

1. 1
Grégoire Nicollier Says:

A shorter proof without solving any equation.
As 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.

2. 2