Find all complex numbers $z$ that satisfy equation $z^3=-8$

Attempt to solve

The real solution is quite easily computable or more specifically complex solution where imaginary part is zero.

$$ z^3=-8 \iff z_1 = \sqrt[3]{-8}=-2 $$

Now WolframAlpha suggests that other complex solutions would be :

$$ z_2 = 1 - i\sqrt{3} $$ $$ z_3 = 1 + i\sqrt{3} $$

Only problem is i don't have clue on how to derive these. I heard something about using polar form of complex number and then increasing argument by $2\pi$ so that we have all roots. I lack intuition on how this would work.

I could try to represent our complex number $-2$ in polar

$$ re^{i \theta} $$

Computing radius via pythagoras theorem

$$ r=\sqrt{(-2)^2+(0)^2} = \sqrt{4}=2 $$

which is quite intuitive even without pythagoras theorem since our imaginary part is $0$ so "radius" has to be same as real part, just without the $-$ sign.

our angle would be $\pi$ radians since our complex number was $-2+i \cdot 0$.

We get:

$$ 2e^{i\pi} $$

Now increasing by every $2\pi$

$$ 2e^{i3\pi},2e^{i5\pi},2e^{i7\pi},\dots $$

which doesn't make any sense since we end up in the same spot over and over again since $2\pi$ in radians is full circle by definition.

  • 1
    $\begingroup$ You are very close with the $re^{i\theta}$ idea. $\endgroup$ – Mandelbrot Oct 10 '18 at 11:06
  • $\begingroup$ You have the $r$ value as two but need $e^{i\theta}$ to be $-1$ $\endgroup$ – Mandelbrot Oct 10 '18 at 11:08
  • $\begingroup$ Clearly $\theta=\pi$ would work but it is not the only value $\endgroup$ – Mandelbrot Oct 10 '18 at 11:10
  • $\begingroup$ Hint: $z^3+8=(z+2)(z^2-2z+4)$ $\endgroup$ – JavaMan Oct 10 '18 at 11:13


$(z+2)(z^2-2z +2^2)=0;$


Solve quadratic equation:

$z_{2,3} = \dfrac{2\pm \sqrt{4-(4)2^2}}{2}$;

$z_{2,3}= \dfrac{2\pm i 2√3}{2}.$

  • $\begingroup$ Use \pm for $\pm$. Also \mp for$\mp$. $\endgroup$ – Oscar Lanzi Oct 10 '18 at 11:21
  • 1
    $\begingroup$ Oscar.Thanks, \pm looks much better:) $\endgroup$ – Peter Szilas Oct 10 '18 at 11:22

We know that because of the $^3$ that there will be three roots. You have rightly determined that the modulus of the roots is $2$. The trick when using polar coordinates is to add $\frac{2\pi}{x}$, where $x$ is the number of roots we have, to the argument. So in this case you would add $\frac{2\pi}{3}$ to the argument.

This gives $z = 2e^{i\pi}, 2e^{\frac{5\pi}{3}i}, 2e^{\frac{\pi}{3}i}.$


Instead of having $-2$ in polar coordinates, use $-8$: $$-8=8e^{i\pi}$$ Repeatedly adding $2\pi i$ to the exponent and cube rooting gives all the cube roots: $$2e^{i\pi/3}\qquad 2e^{i\pi}\qquad 2e^{5i\pi/3}$$ These can be converted back into Cartesian coordinates as $-2$ and $1\pm\sqrt3i$.


Alright, so we have $z^3=8e^{i(\pi+2\pi k)}$ for each $k\in\mathbb{Z}$. Now to find $z$ we need take the third root from the module and divide the argument by $3$. So we get $z=2e^{i(\frac{\pi}{3}+\frac{2\pi}{3}k)}$. So now compute it for $k\in\{0,1,2\}$, after that the roots will repeat. So for $k=0$ we get:


Now put $k=1$ and $k=2$ to find the other roots.


Whenever we deal with exponentials, the $re^{i\theta}$ form often is more convenient.

We know that the magnitude of $-8$ is $8$, so the magnitude of $z$ is $\sqrt[3]8=2$.

So, we know that $$(2e^{i\theta})^3=-8\to e^{i\cdot3\theta}=-1$$

We know that $e^{i\pi}=-1$, so we look for all $\theta$ such that $3\theta = 2\pi\cdot n+\pi$. We find the answers to be $$\frac\pi3,\pi,\frac5{3\pi}$$so we know that our answers are $$2e^{i\frac\pi3},2e^{i\pi},2e^{i\frac{5\pi}3}$$

The a+bi form of these complex numbers is $$\color{red}{-2,1\pm\sqrt3i}$$

  • $\begingroup$ You mean $5\pi/3$ $\endgroup$ – JavaMan Oct 10 '18 at 11:17
  • $\begingroup$ @JavaMan Thank you for catching that! $\endgroup$ – Don Thousand Oct 10 '18 at 11:19

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