Find the value of $z^n+1/z^n$ if $n$ is arbitrary natural number

Question

If $z$ is a complex number satisfying

$$z + \frac{1}{z} = \sqrt{3}$$

then for arbitrary natural number $n$, determine the value of

$$z^n + \frac{1}{z^n}$$

I have tried it with $n=2,3,4$ but it came to different answers.

Answer 1

Multiplying the whole equation by $z$, and solving the obtained quadratic equation, we get:

$z = e^{i\frac{\pi}{6}}$ or $z = e^{-i\frac{\pi}{6}}$

Whether we are in the first case or the second, it makes no difference.

$$z^n + \frac{1}{z^n} = e^{i\frac{\pi n}{6}} + e^{-i\frac{\pi n}{6}} = 2\cos(n\frac{\pi}{6})$$

Answer 2

$$\left(z+z^{-1}\right)\left(z^n+z^{-n}\right)=z^{n+1}+z^{n-1}+z^{-n+1}+z^{-n-1}$$

This can be written as the recurrence relation $$z_{n+1}=\sqrt3z_n-z_{n-1}.$$

The terms are $$2,\sqrt3,1,0,-1,-\sqrt3,-2,-\sqrt3,-1,0,1,\sqrt3,2,\sqrt3,\cdots$$ periodically.

Answer 3

You could solve for $z$. Multiply through by $z$ to transform your equation into a quadratic. You'll get $$z^2-z\sqrt{3}+1=0 \implies z = \frac{\sqrt{3}\pm i }{2}$$ Make sure you check each solution to make sure it works in your original equation. It may help to convert it to polar coordinates, so you can use de Moivre's law.