What is the principal cubic root of $-8$?

Question

According to my book it should be a real number, and according to WolframAlpha it should be $1+1.73i$

What is the correct answer?

Answer 1

The cubic roots of $-8$ are: $$ 2e^{i\pi/3}=2(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}) \qquad 2e^{i\pi}=-2 \qquad 2e^{i2\pi/3}=2(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}) $$

it seems that WolframAlpha consider as principal root the root with the minimum value of the argument, i.e $e^{i\pi/3}$, but usually, if there is a real root this is considered the principal root.

Answer 2

Mathematicians (as opposed to elementary textbook writers) consider complex numbers. To make the principal value of the cube root continuous in as large a region of the complex plane as possible, and to have it agree with the conventional cube root for positive numbers, they choose the cube root with argument in $(-\pi/3 , \pi/3]$. This fits together with a systematic choice of principal value for logarithm, arctangent, and others.

Answer 3

There are three complex cubic roots of $-8$.

Namely $2 e^{i \pi/3}$, $2 e^{i (\pi+2 \pi) /3}$, $2 e^{i (\pi+4 \pi) /3} $. You get this using the polar form(s) of $-8$, that is $8 e^{i \pi}$ and more generally $8 e^{i (\pi+ 2k\pi)}$ for $k $ and integer.

The middle one is $-2$ and the first one is what Wolfram Alpha gives.

The rational for W|A making that choice is that it considers as principal root the one where the principal argument of $-8$, that is $\pi$, is divided by $3$.

The rational of your book should be that they want to be cubic roots of reals to be real.

It is a matter of definition/convention. In some sense I prefer the former, but would use the latter in the right context too.

In any case, you should follow your book or talk to your instructor about it, as otherwise there will be confusion.

Answer 4

In $\mathbb R$, the cube root of $-8$ is $-2$. There is no need for the concept of a "principal root".

So the term "principal cube root" implies that the field of interest is $\mathbb C$, not $\mathbb R$. And then, according to this Wikipedia article, the principal cube root of $z$ is usually defined as $\exp(\frac13\log z)$, where the principal branch of the logarithm is taken. This interpretation gives $1 + i\sqrt 3$ as the principal cube root of $-8$.

By the way, did you notice that Wolfram Alpha has this covered? It lets you choose:

Assuming the principal root | Use the real‐valued root instead

Answer 5

EDIT 1:

I was trying to say about the real part of the complex root. We need be totally lost when helped by the graph of the function. The real root still hangs around an expected location.

If you sketch a graph of $ y = x^3 + 8 $ , it is seen cutting x-axis not only at real x = -2 but the complex roots have real parts coming closest/ farthest to x-axis near about x= 1 when not actually cutting x-axis... depending on lack of harmonicity.

At these points you have complex roots $ 1 \pm i \sqrt {3} $ exactly. $ A view in Complex plane shows all the three positions of vector tip in a WA plot ( not shown here).

To illustrate, I took an irregular graph the red lines point to location of real part of existence of the complex roots.

There are 3 ways non-contacting and intersecting/.. or .. convex/concave $ \cup, \cap $ sort of shaped curves aspected to x-axis in a graph appear e.g., like when $ y=\cosh x, x (1-x), x^6$ respectively. Correspondingly imaginary solutions are odd multiples of $ i \pi/2 $ along with the regular real roots$ (0,1)$ appear..the former non touching cup shapes produce complex roots. The tangential roots are well known.

For $ y(x)= x^3+8 , y^{''}=6x=0 $ a maximum and minimum of y=(x) should occur to the right and left of $x=0$ and complex roots whose real parts should appear nearer to the red abcissas. Exact complex roots can be calculated numerically accurately of course by Newton-Raphson like procedures however.

Answer 6

Consider $x^3 = - 8 $. It "seems" we can just say $x = -2$ because $(- 2)^3 = -8$ but $- 2$ is not the official Primary Solution!

Using De Moivre's theorem again: $$x^3 = - 8$$ $$( r \operatorname{cis}(\theta) )^3 = -8$$ $$r^3 \operatorname{cis}(3\theta) = 8\operatorname{cis}( 180^\circ + 360^\circ n)$$ $$r^3 = 8 \text{ and } 3\theta = 180^\circ + 360^\circ n $$ $$r = 2 \text{ and } \theta = 60^\circ + 120^\circ n = 60^\circ, 180^\circ, 300^\circ $$

$$x_1 = 2\operatorname{cis}( 60^\circ ) = 1 + i\sqrt3$$ $$x_2 = 2\operatorname{cis}(180^\circ) = - 2$$ $$x_3 = 2\operatorname{cis}(240^\circ) = 1 - i\sqrt3$$

The Primary Solution is $x_1 = 1 + i\sqrt3 ≈ 1 + 1.732i$ So the cube root of $-8$ or $(-8)^{\frac13}=1+1.732i$ and NOT $-2$.