Imagine we want to find fourth complex roots of unity of $z=-16$. We first write the number in polar form: $z=16e^{i\cdot\pi}$ Then we use DeMovier theorem to find a fourth root as follow:
$$\begin{align*}
&\sqrt[4]{16e^{i\cdot\pi}}\\
=& \sqrt[4]{16} \cdot (e^{i\cdot\pi})^{\frac{1}{4}}\\
=& 2 \cdot e^{i \cdot \frac{\pi}{4}}
\end{align*}$$
Now I want to find four roots of unity. I know that I can use following formula (again based on DeMovier theorm):
$$
\sqrt[n]{z} = \sqrt[n]{r}\cdot(cos{\frac{2\pi k}{n}} + i\cdot sin{\frac{2\pi k}{n}})
$$
But this would be very time consuming. I am thinking that I can do this more quickly, only by multiplying $2\pi$ on $2 \cdot e^{i \cdot \frac{\pi}{4}}$ four times. But it did not quite works and I do not receive the same result.
Best Answer
So we want to find the $4^{th}$ roots of $z=-16$ $\rightarrow$ we want $z^4=-16$.
Well, $\arg(z)=\pi$ $\Rightarrow z^4 = 16e^{\pi i}$
By De'Moivre we have, $z = (16)^{\frac{1}{4}} (\cos(\frac{\pi+2 \pi k}{4}) + i \sin(\frac{\pi+2 \pi k}{4})) = 2(\cos(\frac{\pi+2 \pi k}{4}) + i \sin(\frac{\pi+2 \pi k}{4}))$, for $k=0,1,2,3$.