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.


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 $k=0,1,2,3$.


The roots are $$ 2\exp \frac{i\pi + 2ik\pi}{4} , k\in \{0,1,2,3\}. $$

In general the $n^{th}$ roots of $r\exp i\theta$ are $$ r^{1/n}\exp \frac{i\theta+ 2ik\pi}n, \ \ k\in \{0,1,\ldots, (n-1)\}. $$

  • $\begingroup$ Is this a general form or just a formula for this problem $e^{i \cdot \frac{pi}{4}}$? $\endgroup$ – bman Mar 30 '14 at 23:04
  • $\begingroup$ it is a general form, replace 0...3 by 0...k when looking at kth root. $\endgroup$ – mookid Mar 30 '14 at 23:06
  • $\begingroup$ In this case instead of $\pi$ we should have $\theta$ because the value of r is different in different cases. So we should have a formula similar to his: $2 exp \frac{\theta + r^{\frac{1}{k}}ik\pi}{4}$. Shouldn't we? $\endgroup$ – bman Mar 30 '14 at 23:08
  • $\begingroup$ exactly. also replace the 2 by $r^{1/k}$ $\endgroup$ – mookid Mar 30 '14 at 23:09
  • $\begingroup$ I am a bit confused. Could you edit the formula in the answer so it suits for a very general form ($r^{i\theta}$)? $\endgroup$ – bman Mar 30 '14 at 23:11

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