Complex solutions of $z^3 + 8 = 0$ (help with resolution method)

Question

How do I solve the equation $z^3 + 8 = 0 *($z$ being a complex number)* using polar coordinates?

These are the first lines of a solution method I saw:

$$z^3 + 8 = 0$$ $$\Longleftrightarrow z^3 =-8=8e^{i\pi+ 2 \pi in}$$ ($n$ is a natural number)

Could you explain why is $-8$ equal to that exponential multiplied by $8$?

Thank you so much for your help

Answer 1

Let $z=re^{i\theta}$

using the famous relation $e^{i\pi}=-1$,

the equation becomes

$$r^3e^{3i\theta}=2^3e^{i\pi +2in\pi}$$

then

$r=2$ and $\theta=\frac{\pi}{3}(1+2n)$. with $n \in \mathbb Z$.

Answer 2

$$ z^3 = -8 $$

In polar coordinates you have $z = r e^{i\theta} = r(\cos\theta+i\sin\theta),$ where $\theta$ is real and $r$ is real and nonnegative.

A basic fact you need here is that $z^n = r^n(\cos(n\theta) + i\sin(n\theta)).$ If $n=3$ that says $z^3 = r^3(\cos(3\theta)+i\sin(3\theta)).$

Thus $-8 = 8\cdot( -1 + (i\cdot0) ),$ so you need $r^3=8$ and $\cos(3\theta)=-1$ and $\sin(3\theta)=0.$

You have $\cos=-1$ and $\sin=0$ at $180^\circ$ or $\pi$ radians. One-third of that is $60^\circ$ or $\pi/3$ radians. At that point you have $$ \cos60^\circ = \frac 1 2 \quad\text{and}\quad\sin60^\circ = \frac{\sqrt 3} 2. $$ Thus you have $z = 2\left(\dfrac 1 2 + i\dfrac{\sqrt3} 2\right) = 1 + i\sqrt 3.$

However, sine and cosine are periodic with a period of $360^\circ$, so $\theta\mapsto(\cos(3\theta)+i\sin(3\theta))$ is periodic with a period of $120^\circ$. Thus adding $360^\circ/3$ or $120^\circ$ gives you another solution: $$ 2(\cos180^\circ+i\sin180^\circ) = 2(-1+0i) = -2. $$ Then add another $120^\circ$ to get a third solution: $$ 2(\cos300^\circ + i\sin300^\circ) = 2\left( \frac 1 2 - i\frac{\sqrt 3} 2 \right) = 1 - i\sqrt3. $$

Another $120^\circ$ brings you back to where you started.

Algebra alone will tell you there cannot be more than three solutions, as follows. Suppose you find one solution to $z^3+8=0$. Call it $a$. Then $$ z^3 + 8 = (z-a)(\cdots\cdots\cdots). $$ The other factor besides $(z-a)$ can be found by long division, and must be a $2$nd-degree polyonomial. Then we find another solution -- call it $b$. Then we have $z^3+8 = (z-a)(z-b)(\cdots\cdots).$ A third solution gives us $(z-a)(z-b)(z-c)(\cdots\cdots).$ But the last factor must be constant since otherwise we'd go up to higher than $3$rd degree.

(The specifics: Once you know $z=-2$ is a solution, you have $$ z^3+8 = (z-2)(\cdots\cdots\cdots). $$ Long division yields $$ z^3+8 = (z-2)(z^2+2z+4). $$ Then solve $z^2+2z+4=0$ by the usual methods used for quadratic equations and get $z = 1 \pm i\sqrt 3,$ agreeing with what we got by polar coordinates.