Find all positive integers $n$ such that $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$?

Question

Find all positive integers $n$ such that $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$ ?

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I wrote $\displaystyle \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n = 2$ as $$2\Re\left(\left({-1 + i\sqrt{3}\over 2}\right)^n\right) = 2 \implies \Re\left(\left({-1 + i\sqrt{3}\over 2}\right)^n\right) = 1$$

But I don't know how to proceed now. I can't use polar representation of complex numbers. Please provide me some direction.

Answer 1

First of all, note that $\left| \frac{-1 \pm i\sqrt 3}{2} \right| = 1$, hence this holds for all $n$th powers of these quantities also.

The triangle inequality states that $|a| + |b| \geq |a+b|$, with equality holding if and only if $a$ is a scalar multiple of $b$.

In our case, if we assume that the final condition holds, we have that for $a,b = \left(\frac{-1 \pm i\sqrt 3}{2}\right)^n$ (it doesn't matter which is which), $2 = |a| + |b| \geq |a+b| = 2$, which means equality is being obtained in the triangle inequality.

This means that $\left(\frac{-1 + i\sqrt 3}{2}\right)^n$ is a scalar multiple of $\left(\frac{-1 - i\sqrt 3}{2}\right)^n$. However, since both have modulus $1$, we can check that this happens if and only if $\left(\frac{-1 + i\sqrt 3}{2}\right)^n = \left(\frac{-1 - i\sqrt 3}{2}\right)^n$.

Hence, dividing both sides by $\left(\frac{-1 - i\sqrt 3}{2}\right)^n$ (it can't be zero, as it has modulus $1$), we get that (I leave you to see this) $\left(\frac{-1 + i\sqrt{3}}{2}\right)^n = 1$.

If you know your roots of unity well, this gives $n = 3k$, where $k$ is an integer.

Answer 2

We have $\dfrac{-1+\sqrt3i}2=e^{i120^\circ}, \dfrac{-1-\sqrt3i}2=e^{-i120^\circ}$

$$2= \left({-1 + i\sqrt{3}\over 2}\right)^n + \left({-1 - i\sqrt{3}\over 2}\right)^n=e^{i120^\circ n}+e^{-i120^\circ n}=2\cos(120^\circ n) $$

$$\iff\cos(120^\circ n)=1$$

$$\implies120^\circ n=360^\circ m$$ where $m$ is any integer

Answer 3

When you get stuck on a problem like this, it often helps to simply compute some examples, in hopes you'll spot a pattern in the results. In this case

$$\left(-1+i\sqrt3\over2\right)^2=\left(1-2i\sqrt3-3\over4\right)=\left(-2-2i\sqrt3\over4\right)=\left(-1-i\sqrt3\over2\right)$$

so

$$\left(-1+i\sqrt3\over2\right)^3=\left(-1+i\sqrt3\over2\right)\left(-1+i\sqrt3\over2\right)^2=\left(-1+i\sqrt3\over2\right)\left(-1-i\sqrt3\over2\right) =\left(1+3\over4\right)=1$$

at which point you might well say Wow! If you don't, just keep going until you do:

$$\left(-1+i\sqrt3\over2\right)^4=\left(-1+i\sqrt3\over2\right)\left(-1+i\sqrt3\over2\right)^3=\left(-1+i\sqrt3\over2\right)\cdot1=\left(-1+i\sqrt3\over2\right)$$

$$\left(-1+i\sqrt3\over2\right)^5=\left(-1+i\sqrt3\over2\right)^2\left(-1+i\sqrt3\over2\right)^3=\left(-1-i\sqrt3\over2\right)\cdot1=\left(-1-i\sqrt3\over2\right)$$

$$\left(-1+i\sqrt3\over2\right)^6=\left(-1+i\sqrt3\over2\right)^3\left(-1+i\sqrt3\over2\right)^3=1\cdot1=1$$

Once you see the pattern, then you can hope to understand it along the lines of the other answers that have already been posted.

In short, there is no need to wait for a burst of inspiration that solves a problem in one fell swoop; playing with examples is often illuminating, and can pave the way to understanding. At the very least it gives you something to do while you're puzzling over how to approach the probem.