Why $8^{\frac{1}{3}}$ is $1$, $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$

Question

The question is:

Use DeMoivre’s theorem to find $8^{\frac{1}{3}}$. Express your answer in complex form.

Select one:

a. 2

b. 2, 2 cis (2$\pi$/3), 2 cis (4$\pi$/3)

c. 2, 2 cis ($\pi$/3)

d. 2 cis ($\pi$/3), 2 cis ($\pi$/3)

e. None of these

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I think that $8^{\frac{1}{3}}$ is $(8+i0)^{\frac{1}{3}}$

And, $r = 8$

And, $8\cos \theta = 8$ and $\theta = 0$.

So, $8^{\frac{1}{3}}\operatorname{cis} 0^\circ = 2\times (1+0)=2$

I just got only $2$. Where and how others $\frac{2\pi}{3}$, and $\frac{4\pi}{3}$ come from?

Answer 1

We could look at it like this:
$$8^{\frac13}=2.1^{\frac13}=2\cdot \text{CiS}\left(\frac{2k\pi}{n}\right)$$ Now for different values of $k$, we have different answers: (here $n$ is $3$) $$k=1\implies 8^{\frac13}=2\cdot\text{CiS} \left(\frac{2\pi}{3}\right)$$ $$k=2\implies8^{\frac13}=2\cdot\text{CiS}\left(\frac{4\pi}{3}\right)$$ $$k=3\implies8^{\frac13}=2\cdot\text{CiS}(2\pi)=2$$

You could read up on $n^{\text{th}}$ roots of unity on Wikipedia to get a better picture

Answer 2

Here, $$\begin{align*} 8^{1/3} &= (|8|e^{2\pi kj})^\frac{1}{3}, k = 0,1,2\\ &= |8|^\frac{1}{3} e^{\frac{2}{3}\pi kj}, k = 0,1,2\\ &= 2 e^{\frac{2}{3}\pi kj}, k = 0,1,2\\ \end{align*}$$

so,for $k=1$,$k=2$ we get $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$

Or take: $$8^{1/3}=x$$ Then we get,

$$(x-2)(x^2+2x+4)=0$$

Then we get our desired roots.

Answer 3

$8^{\frac{1}{3}}$=$2(1)^{\frac{1}{3}}=2,2\omega,2{\omega}^2$

here $\omega$ is cube root of unity

Answer 4

Let $z^3=8$.

Thus, $$(z-2)(z^2+2z+4)=0,$$ which gives $$\{2,-1+\sqrt3i,-1-\sqrt3i\}$$ or $$\left\{2(\cos0+i\sin0),2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right), 2\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)\right\}$$