Compute all roots of $(-8)^{\frac{1}{3}}$

Question

$$(-8)^{\frac{1}{3}}$$

The problem states to compute all roots of the complex number above. Below is my attempt, but my inquiries are if I did it right and why it doesn't match Wolfram. Wolfram only returns $1$ result - was I only supposed to have $1$ as well??

$$\text{modulus} = 8$$ $$\theta = \arctan \left(\frac{0}{-8}\right) = 0$$

Because we are working in the left half of the complex plane, $\theta=\pi$.

$$z=\left(8e^{i(2\pi k+\pi)}\right)^{\frac{1}{3}}$$

This complex number has $3$ roots, so $k=0,1,2$ :

$$z_0=2e^{i\frac{\pi}{3}}$$ $$z_1=2e^{i\pi}$$ $$z_2=2e^{i\frac{5\pi}{3}}$$

Answer 1

You want to solve:

$$z^3=-8\Longleftrightarrow$$ $$z^3=|-8|e^{\arg(-8)i}\Longleftrightarrow$$ $$z^3=8e^{\pi i}\Longleftrightarrow$$ $$z=\left(8e^{\left(2\pi k+\pi\right)i}\right)^{\frac{1}{3}}\Longleftrightarrow$$ $$z=2e^{\frac{1}{3}\left(2\pi k+\pi\right)i}$$

With $k\in\mathbb{Z}$ and $k:0-2$

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So the solutions are:

$$z_0=2e^{\frac{1}{3}\left(2\pi\cdot0+\pi\right)i}=2e^{\frac{\pi i}{3}}$$ $$z_1=2e^{\frac{1}{3}\left(2\pi\cdot1+\pi\right)i}=2e^{\pi i}=-2$$ $$z_2=2e^{\frac{1}{3}\left(2\pi\cdot2+\pi\right)i}=2e^{-\frac{\pi i}{3}}$$

Answer 2

We can use elementary factorization to find all roots. Let $x = \left(-8\right)^{\frac{1}{3}}\Rightarrow x^3 = -8\Rightarrow x^3 + 8 = 0\Rightarrow (x+2)(x^2-2x+4) = 0\Rightarrow (x+2)((x-1)^2 + 3) = 0\Rightarrow x = -2, (x-1)^2 = -3\Rightarrow x = -2, x-1 = \pm i\sqrt{3}\Rightarrow x = -2, 1 \pm i\sqrt{3}$.

Answer 3

The problem you are solving and the problem Wolfram Alpha is solving are different problems.

You want to find all of the roots of $x^3+8=0$ which you did corrrectly. Your three solutions are all of the solutions to that equation.

However, you asked Wolfram Alpha to take $-8$ to the $1/3$ power. That's a different problem because there, you're not solving a polynomial equation. Instead, you're taking one number to the power of a fractional number. When you take a number to the power of another number, there is only one clearly defined answer even if the power is fractional. For example, $2^{\frac 1 2}$ is $\sqrt 2$, the principal (positive) square root of 2, and that's different from $-\sqrt 2$. Similarly, $(-8)^{\frac 1 3}$ is the "principal" cube root of $-8$ and must be $2e^{\frac{i\pi}{3}}$ because out of all of the solutions to $x^3+8=0$, that solution has the lowest argument.

If you want to find all of the solutions to $x^3+8=0$, tell Wolfram Alpha to do that instead. As you can see, Wolfram Alpha now has three answers like you do, but has expressed them in rectangular form rather than polar form like you did. However, your answers and Wolfram Alpha's answers are still the same.

Answer 4

It “seems” we can just say x = –2 because (– 2)^3 = –8 but – 2 is not the Primary or Principal Solution!

Using De Moivre’s theorem again: x^3 = – 8 ( r cis(θ) )^3 = –8 r^3 cis(3θ) = 8cis( 180 + 360n) r^3 = 8 and 3θ = 180 + 360n r = 2 and θ = 60 + 120n = 60, 180, 300

x1 = 2cis( 60 ) = 1 + i√3 x2 = 2cis(180) = – 2 x3 = 2cis(300) = 1 – i√3

The Primary Solution is x1 = 1 + i√3 ≈ 1 + 1.732i So cube root of -8 or (-8)^(1/3)=1+1.732i and NOT-2!