What about the solutions of $z^{1/3} +1 = 0$?

Question

I'm trying to find the zeros of the equation $$z^{1/3} +1 = 0.$$

My professor said that the solutions are the third roots of unity multiplied by $-1$. My problem is that when I calculate the cubic root of one of the numbers $$\bigg\{e^{i \pi},e^{i \pi/3},e^{-i \pi/3}\bigg \},$$ in order to verify that these numbers are really the numbers that give me $z^{1/3}+1=0$, I obtain one of the following sets: $$(e^{i \pi})^{1/3} = \bigg\{e^{i \pi},e^{i \pi/3},e^{-i \pi/3}\bigg \},$$ $$(e^{i \pi/3})^{1/3} = \bigg \{ e^{i \pi/9},e^{7 i \pi/9},e^{-5 i \pi/9} \bigg \},$$ $$(e^{-i \pi/3})^{1/3} = \bigg \{ e^{-i \pi/9},e^{-7 i \pi/9},e^{5 i \pi/9} \bigg \}.$$

First of all, if I consider the sum of a complex and a set element-wise, only one of the sets gives me $0$ when one is added to it (it is $(e^{i \pi})^{1/3}.$)

If the sum of a set and a complex number isn't element-wise, what means, for example, $(e^{i \pi})^{1/3} + 1 = 0$ (supposing it is a root as my professor said)? We are comparing a set with a number, must be interpreted $0$ as the set $\{0\}$?

Furthermore, if I interpret $0$ as a set, I don't have the equality of the sets, and for $(e^{i \pi/3})^{1/3}, (e^{-i \pi/3})^{1/3}$ I have that $\{0\}$ is not a subset of $(e^{i \pi/3})^{1/3}+1, (e^{-i \pi/3})^{1/3}+1$, respectively.

Note: When I'm considering the cubic root of the solutions proposed is only in order to see that these are roots really, and then I get stuck since the complex cubic root is a multivalued function.

Thanks to everyone!

Answer 1

It seems you are looking for something different that is

$$z^{3} +1 = 0 \iff z^3=-1 $$

and the suggestion by your professor is simply to evaluate $w^3=1$ and then obtain the solution from here using that

$$w_i^3=1 \implies (-1\cdot w_i)^3=-1\cdot (w_i)^3=-1$$

Answer 2

$$z^{1/3}+1=0\implies z=(-1)^3$$

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$$z^3+1=0\implies z=(-1)^{1/3}=-1^{1/3}=-(e^{i2k\pi})^{1/3}$$

and

$$z=-1\lor-\cos\frac{2\pi}3-i\sin\frac{2\pi}3\lor-\cos\frac{2\pi}3+i\sin\frac{2\pi}3.$$

Answer 3

If you want to explicitly account for the possibility that $-1$ may not be the only solution to $z^{1 / 3}=-1$ you could proceed as follows:

$-1 = e^{(2n+1)\pi} \space n\in \mathbb{Z}$

$\Rightarrow z = (z^{1/3})^3 = e^{(6n+3)\pi} \space n\in \mathbb{Z}$

but $6n+3 = 2m+1$ where $m=3n+1$

$\Rightarrow z = e^{(6n+3)\pi} = e^{(2m+1)\pi} = -1 \space \forall n \in \mathbb{Z}$