Complex Numbers Problem Solving 1

$x = e^{i\frac{\pi}{3}}$

$y = e^{i\frac{5\pi}{3}}$

$\Rightarrow x^2 = e^{i\frac{2\pi}{3}}$

$\Rightarrow y^2 = e^{i\frac{10\pi}{3}}$

Thus, we need to find,

$\left(\frac{e^{i\frac{2\pi}{3}}}{e^{i\frac{5\pi}{3}}}+\frac{e^{i\frac{10\pi}{3}}}{e^{i\frac{\pi}{3}}}\right)^6$

$\left(e^{-i\pi}+e^{3\pi}\right)^6 = \left(-1+(-1)\right)^6 = \left(-2\right)^6 = \boxed{64}$

Note that

$(\frac{x^2}{y} + \frac{y^2}{x} )^6 = (\frac{x^3+y^3}{xy} )^6$

We can use the De Moivre's Theorem which states that

If $z$ is a complex number of the form $z = r(\cos \theta + i \sin \theta)$ then for any positive integer $n$ , $z^n = r^n (\cos n\theta + i \sin n\theta)$

Also,

if $z_1 = r_1 (\cos \theta_1 + i \sin \theta_1)$ and $z_2 = r_2 (\cos \theta_2 + i \sin \theta_2)$ are two complex numbers, then $z_1 \times z_2 = r_1 \times r_2 [\cos (\theta_1 +\theta_2) + i \sin (\theta_1 +\theta_2)]$

Thus,

$x^3 = [\cos (\frac{\pi}{3}) + i \sin (\frac{\pi}{3})]^3= \cos 3(\frac{\pi}{3}) + i \sin 3(\frac{\pi}{3}) = \cos\pi + i \sin \pi = -1 + 0i = -1$

and $y^3 =[\cos (\frac{5\pi}{3}) + i \sin (\frac{5\pi}{3})]^3 =\cos 3(\frac{5\pi}{3}) + i \sin 3(\frac{5\pi}{3}) = \cos 5\pi + i\sin 5 \pi = -1 + 0i = -1$

On the other hand,

$xy = (\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) \times (\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3}) = \cos 2\pi + i \sin 2\pi = 1+ 0i = 1$

Finally,

$(\frac{x^2}{y} + \frac{y^2}{x} )^6 = (\frac{x^3+y^3}{xy} )^6 = (\frac{-1 + (-1) }{1})^6 = (\frac{-2}{1})^6 = (-2)^6 = 64$