So i got this problem :
Find all the roots of $r^{3}=(-1)$
i can only think to use :
$\sqrt[n]{z} =\sqrt[n]{r}\left[\cos \left(\dfrac{\theta + 2\pi{k}}{n}\right) + i \sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right] $
i know that $\theta = \tan^{-1}(\dfrac{b}{a}) $as in from $ z=a + bi$
$r=\sqrt[]{a^2 + b^2}$ , $k=0,1,2,3,4,.. n-1$
so anyone could explain this to me ? lets say that $r^{3}=(-1) == z^{3}=(-1)$ a since i dont like it and it messes me up using the formula
so far :
$ z^{3} = (-1) $ so $z = \sqrt[3]{(-1 + 0 \cdot i)}$
$ r= \sqrt[]{(-1)^{2} + 0^{2}}=1$
$\tan^{-1}(\dfrac{b}{a})=\tan^{-1}(\dfrac{0}{-1})=\tan^{-1}(0)=0$ <- not sure about his part
$ z^{3}=\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 0}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 0}{3})$
$ =\cos(\dfrac{\pi}{3}) +i\cdot \sin(\dfrac{\pi}{3})= \dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2}$

$ \sqrt[3]{1}(\cos(\dfrac{\pi+2\pi \cdot 1}{3})+i \cdot \sin(\dfrac{\pi+2\pi \cdot 1}{3})$
$=\cos(\pi) + i \cdot \sin(\pi)=(-1) + 0 = -1$

$\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 2}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 2}{3})$
$=\cos(\dfrac{5\pi}{3})+i \cdot \sin(\dfrac{5\pi}{3})=-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2}$
so is this correctet or im missing something here? Roots $(\dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2};-1;-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2})$

  • $\begingroup$ Since $|r^3|=|-1|=1$ it follows that $|r|=1$ hence our roots lie on the unit circle in the complex plane. In fact, since $r^3=-1$ we know that our roots are precisely those complex numbers that partition said unit circle into three equal parts; one such number is obviously $-1$. Can you determine the others? en.wikipedia.org/wiki/Root_of_unity $\endgroup$
    – obataku
    Aug 29, 2013 at 12:31
  • $\begingroup$ As you suspected, the angle is incorrect. The $\theta$ for $-1$ is $\pi$, not $0$. The problem shows up because $\tan$ is not injective, so we can't make a nice inverse. For all $x$, there are two $\theta \in [0, 2\pi)$ such that $\tan(\theta) = x$, and they lie on opposite sides of the unit circle. (Can you see why?). Your $==$ notation makes me think you're a programmer, so it may help to look up functions called "atan2(x,y)". $\endgroup$ Aug 30, 2013 at 3:14

5 Answers 5


As an alternative to finding "roots of unity" using the "$\operatorname{cis} \theta$" approach, you can easily solve the equation $$r^3 = -1 \iff r^3 + 1 = 0 \iff (r+1)(r^2 - r + 1)$$

That gives a real root $$r_1 = -1,$$ and two complex roots, which can be solved by applying the quadratic formula $$\left(\dfrac {-b \pm \sqrt {b^2 - 4ac}}{2}\right)\;\;\text{to}\;\;r^2 -r + 1 = 0\quad\text{to find}\;\;r_2, r_3$$

$$r_2, r_3 = \frac 12 \pm \frac{\sqrt {-3}}{2} = \frac 12 \pm \frac{\sqrt 3}2\;i$$

  • $\begingroup$ This needs a TU. +1 $\endgroup$
    – Amzoti
    Aug 29, 2013 at 12:17
  • $\begingroup$ @amWhy $\Large{+}^+$ ;) $\endgroup$
    – Software
    Aug 31, 2013 at 13:41


Using $\,cis(\theta):=\cos\theta+i\sin\theta=cis(\theta+2k\pi)\;,\;\;k\in\Bbb Z\;$ , we have:

$$z=r\,cis(\theta)\implies -1=cis(\pi+2k\pi)=z^3=r^3\,cis(3\theta)\;\;\text{(de Moivre's Theorem)}\implies$$

$$z_k:=r\,cis\left(\frac\pi3(1+2k)\right)\;,\;\;k=0,1,2\;\;\text{(why is it enough to take these values of}\;\;k)?$$

Now calculate the roots, taking into account that $\,r=1\;$ (why?)

  • 1
    $\begingroup$ +1 Good point of remarking $cis$. This concise form is used here hardly. $\endgroup$
    – Mikasa
    Aug 30, 2013 at 8:59

Looking as an extension of fields, $x^3+1=0$ has exactly 3 unique solutions. $(x-\alpha)(x-\beta)(x+1)=x^3+1=(x+1)(x^2-x+1)$ (-1 is apparent.) You can solve for the other 2 by binomial or trig or whichever way you want. but as far as the trig goes, $cos(\theta)+i*sin(\theta) = e^{i\theta}$, so $(r*cos(\theta)+i*r*sin(\theta))^k = r^ke^{ik\theta} =r^kcos(k\theta)+i*r^ksin(k\theta)$ So it is just a matter of setting $r^3cos(3\theta)+i*r^3sin(3\theta)=-1$ where r is a positive number.


Hint: An alternative to $ -1=-1+0\cdot i=\cos(-\pi)+i\sin(-\pi)\quad $ one ungainly brute force $$ (a+bi)^3= -1+0\cdot i \Longleftrightarrow a^3-3ab^2=-1 \mbox{ and } -b^3+3a^2b=0 $$


Just to hop and skip around some of the theory hovering in the background of some of these answers...

O.K., firstly by the Fundamental Theorem of Algebra $r^3=-1$ has at most three (distinct) solutions. This means that if you can find three then you are done (this is why in the '$k$' solutions they only need $k=0,1,2$).

Obviously $-1$ is one solution. Now what we want to do is find another (complex) number that when cubed gives you $-1$. Now how do complex numbers multiply? What does it look like? Multiplication for complex numbers is better handled when we write in polar form

$$z=|z|(\cos \theta+i\sin\theta),$$

where $|z|$, the magnitude, is the distance from $z$ to the origin and $\theta$, the argument $\arg(z)$, is the\an angle between the positive $x$-axis and the ray connecting $z$ to the origin.

It turns out that when you multiply complex numbers together they stretch\contract and rotate... more precisely

$$|z_1\cdot z_2|=|z_1||z_2|\text{ and }\arg(z_1\cdot z_2)=\arg(z_1)+\arg(z_2).$$

You can show this by multiplying together

$$z_1\cdot z_2=\left(|z_1|(\cos \theta_1+i\sin\theta_1)\right)\cdot\left(|z_2|(\cos \theta_2+i\sin\theta_2)\right).$$

Now $|-1|=1$ so we want $|r^3|=|r|^3=1$ and because $|r|\geq0$ we need $|r|=1$.

Next $\arg(-1)=\pi$ and $\arg(r^3)=3\arg(r)$ so we want

$$\pi=3\arg(r)\Rightarrow \arg(r)=\pi/3,$$

and indeed by DeMoivre (AKA multiplying complex numbers by themselves)

$$(\cos(\pi/3)+i\sin(\pi/3))^3=\cos \pi+i\sin \pi=-1$$

as required.

Now that is two solutions. Finally by the Conjugate Root Theorem, as the coefficients of $r^3=-1$ are all real the conjugate of this solution is also a root.

So we have the solution set



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