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

## Problem

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

$$re^{i \theta}$$

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

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.

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

$$z^3+8=0;$$

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

$$z_1=-2;$$

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

$$z_{2,3}= \dfrac{2\pm i 2√3}{2}.$$

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

$$z_1=2e^{i\frac{\pi}{3}}=2(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}))=1+i\sqrt{3}$$

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 $$\color{red}{-2,1\pm\sqrt3i}$$

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