Complex numbers - how to solve $(\sqrt{3}-i)z^6+16=0$ When $z = x + yi$ (or $a + bi$), I need to solve:
$$(\sqrt{3}-i)z^6+16=0$$
Here is how I started:
$(\sqrt{3}-i)z^6=-16$
$z^6=\frac{-16}{\sqrt{3}-i}$
$z=\sqrt[6]\frac{-16}{\sqrt{3}-i}$
In other cases I get a normal complex number under the root sign in the right side (in the form of $x+yi$) and then I represent this number in its trigonometric form and apply De Moivre's formula. But this case seems different... What should I do next?
 A: $$\frac{-16}{\sqrt3-i}=\frac{(-16)}{(\sqrt3-i)}.\frac{(\sqrt3+i)}{(\sqrt3+i)}=\frac{-16\sqrt3-16i}{4}=-4\sqrt3-4i$$
Now by $x=r\cos\theta;y=r\sin\theta$ and $r=\sqrt{x^2+y^2}$ 
$$r=\sqrt{48+16}=8$$
So $$-4\sqrt3=8\cos\theta$$
$$\frac{-\sqrt3}{2}=\cos\theta$$
So $\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$

So we get that $$-4\sqrt3-4i=8(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$$
Now use De Moivre's Theorem or Euler's Formula as per your comfort. 
$$\text{De Movire's Theorem}$$
$$(\cos\theta+i\sin\theta)^{\frac1n}=(\cos\frac{\theta}{n}+i\sin\frac{\theta}{n})$$ 
$$\text{Euler's Formula}$$
$$e^{i\theta}=\cos\theta+i\sin\theta$$
A: $$(\sqrt{3}-i)z^6+16 = 0 \Longleftrightarrow$$
$$(\sqrt{3}-i)z^6 = -16 \Longleftrightarrow$$
$$z^6 = \frac{-16}{\sqrt{3}-i} \Longleftrightarrow$$
$$z^6 = -4\sqrt{3}-4i \Longleftrightarrow$$
$$z^6 = 8e^{-\frac{5\pi}{6}i} \Longleftrightarrow$$
$$z^6 = 8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \left(8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i}\right)^{\frac{1}{6}} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{1}{6}\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{1}{6}\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{(-5+12k)\pi}{36}i}$$
With $k\in\mathbb{z}$ and $k=0-5$
With Eulers formula we found:
$$z = \sqrt{2}e^{\frac{(-5+12k)\pi}{36}i}=\sqrt{2}\left(\cos\left(\frac{(-5+12k)\pi}{36}\right)+\sin\left(\frac{(-5+12k)\pi}{36}\right)i\right)$$
