Question

Determinati n∈Z pentru care (1+i√3)ⁿ + (1-i√3)ⁿ = 2ⁿ

Answer 1

$(1+i\sqrt{3})^n+(1-i\sqrt{3})^n = 2^n \Big|:2^n$

$\Rightarrow \left(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\right)^n+\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)^n = 1$

$\Rightarrow \left(\cos\frac{\pi}{3}+i\sin \frac{\pi}{3}\right)^n+\left(\cos \frac{-\pi}{3}+i\sin \frac{-\pi}{3}\right)^n = 1$

$\Rightarrow \left(\cos \frac{n\pi}{3}+i\sin \frac{n\pi}{3}\right) + \left(\cos \frac{-n\pi}{3}+i\sin \frac{-n\pi}{3}\right) = 1$

$\Rightarrow \cos \frac{n\pi}{3}+i\sin \frac{n\pi}{3} + \cos \frac{n\pi}{3}-i\sin \frac{n\pi}{3} = 1$

$\Rightarrow 2\cos \frac{n\pi}{3} = 1$

$\Rightarrow \cos\frac{n\pi}{3} = \frac{1}{2}$

$\Rightarrow \dfrac{n\pi}{3} = \pm \arccos\frac{1}{2}+2k\pi$

$\Rightarrow \dfrac{n\pi}{3} = \pm \dfrac{\pi}{3}+2k\pi\Big|\cdot \dfrac{3}{\pi}$

$\Rightarrow \boxed{n = \pm 1+6k,\,\,\,\, k\in \mathbb{Z}}$