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

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!

• There must be a confusion in the exponents. The solution of the given equation is $z=(-1)^3=-1$. And why do you take the cube roots of the solutions ??? – Yves Daoust Dec 10 '18 at 8:10
• "My problem is that when I calculate the cubic root of one of the numbers {eiπ,eiπ/3,e−iπ/3}" Why are you trying to find the cubic roots of those numbers? The problem asks you to find the cubic roots of $-1$ and your professor says to try to find the cubic roots of $1$ first and multiple by $-1$. So why are you trying to find the cubic roots of those numbers? – fleablood Dec 10 '18 at 8:13
• @fleablood: also note that the equation does not require cubic roots at all... – Yves Daoust Dec 10 '18 at 8:17
• @YvesDaoust I take the cubic roots in order to verify the solutions, and if the cubic root of them is equal to -1. I already know that he is wrong about the solutions but -1 is a good aspirant. But I don't know how to interpret the equalities between these sets when I'm verifying the -1 as a solution. – AlgebraicallyClosed Dec 10 '18 at 8:20
• @AlgebraicallyClosed: read my comment thoroughly and fix ! – Yves Daoust Dec 10 '18 at 8:22

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

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

• Sorry if I am misunderstanding you, but I know how to solve the equation z^3+1=0, my problem is that the multivaluedness of (·)^(1/3) confuses me and when I try to verify the solution z=-1 it takes three values and only one of them is equal to -1. Sorry if I am not following correctly your answer. Anyway, thanks for your answer! – AlgebraicallyClosed Dec 10 '18 at 8:34
• @AlgebraicallyClosed What I mean is the it seems you are facing a different problem to that your professor is asking for. The suggestion given seems indicate the resolution for $z^{3} +1 = 0$ and not $z^{1/3} +1 = 0$. – gimusi Dec 10 '18 at 8:42
• @AlgebraicallyClosed Note moreover that $z^{1/3} +1 = 0 \implies z=-1$ which is the only solution. – gimusi Dec 10 '18 at 8:44
• I know that the only possible solution is -1. Could you help me in verifying that it's really a solution? What I say is "Could you do the steps that verify that it is really a solution?'' I think I'm misunderstanding everyone who is answering. The point of view I'm taking is that according to the branch we fix to work in, I think that exists or not exists a solution, and it's a bit confusing to me talk about solutions of the equation without this prerequisites, and this way my professor solved the problem in class, without talking of branches. – AlgebraicallyClosed Dec 10 '18 at 9:29
• The check is $(-1)^{\frac13}=-1$ indeed $(-1)^3 = -1$. – gimusi Dec 10 '18 at 9:31

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 $$m=3n+1$$

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

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

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

and

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

• Note that the teacher's advice is not so useful, as you can work directly with the roots of $-1$. – Yves Daoust Dec 10 '18 at 8:29
• Sorry, but I still have the problem of the multivaluedness of (·)^(1/3), because when I try to verify that z=-1 is a solution, I have that (-1)^(1/3) = {-1,1/2-sqrt(3)/2i,1/2+sqrt(3)/2i}, and only one of these values give me zero when is added to 1. I'm sorry for my awkwardness and thaks anyway. – AlgebraicallyClosed Dec 10 '18 at 8:39
• @AlgebraicallyClosed: sorry, fix your question first. – Yves Daoust Dec 10 '18 at 8:42
• I have rewritten my question, I hope it's now understood. – AlgebraicallyClosed Dec 10 '18 at 8:55
• @AlgebraicallyClosed: you are still missing the point completely. Read all comments of everyone ! – Yves Daoust Dec 10 '18 at 8:57