<s>[INST] 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?
 [/INST] $$(\sqrt{3}-i)z^6+16 = 0 \Longleftrightarrow$$
$$(\sqrt{3}-i)z^6 = -16 \Longleftrightarrow$$
$$z^6 = \frac{-16}{\sqrt{3}-i} \Longleftrightarrow$$
$$z^6 = -4\sqrt{3}-4i \Longleftrightarrow$$
$$z^6 = 8e^{-\frac{5\pi}{6}i} \Longleftrightarrow$$
$$z^6 = 8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \left(8e^{\left(-\frac{5\pi}{6}+2\pi k\right)i}\right)^{\frac{1}{6}} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{1}{6}\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{1}{6}\left(-\frac{5\pi}{6}+2\pi k\right)i} \Longleftrightarrow$$
$$z = \sqrt{2}e^{\frac{(-5+12k)\pi}{36}i}$$
With $k\in\mathbb{z}$ and $k=0-5$
With Eulers formula we found:
$$z = \sqrt{2}e^{\frac{(-5+12k)\pi}{36}i}=\sqrt{2}\left(\cos\left(\frac{(-5+12k)\pi}{36}\right)+\sin\left(\frac{(-5+12k)\pi}{36}\right)i\right)$$
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