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I have seemingly very straight forward question, find all the roots of $8^{1/3}$. However, I feel that I am skipping a key step or something is just not correct.

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

Essentially, at this point, plug in $k$ yields all roots.

However, what is the fundamental difference between $8^{1/3}$ and $|8|^{1/3}$?

It seems that I am cheating by writing $|8|^{1/3} = 2$. Isn't it true that $|8|^{1/3} = 8^{1/3}$ (since $8$ is a positive real number)?

How can I rewrite the above set of equations so I don't have this kind of confusion.

José Carlos Santos
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Olórin
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3 Answers3

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In fact, since $\lvert8\rvert=8$, $\lvert8\rvert^\frac13=8^\frac13$.

If $z^3=8$, then $z\neq0$ and therefore $z=re^{\theta i}$ for some $r>0$ and some real $\theta$. But then$$z^3=8\iff r^3e^{3\theta i}=8\iff r^3=8\wedge e^{3\theta i}=1.$$Now, there's only one number $r>0$ whose cube is $8$: $r=2$. On the other hand,$$e^{3\theta i}=1\iff3\theta i=2k\pi i$$for some integer $k$. But then $\theta=\frac{2k\pi}3$. So, there are $3$ and only $3$ solutions: $2$ (take $k=0$), $-1+\sqrt3\,i$ (take $k=1$) and $-2-\sqrt3\,i$ (take $k=2$). If you use any other integer $k$, you will get again one of these.

José Carlos Santos
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Hint: with $$8^{1/3}=x$$ we get$$(x-2)(x^2+2x+4)=0$$

Dr. Sonnhard Graubner
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There are no difference between $8^{1/3}$ and $|8|^{1/3}$, $|8|$ means, in complex plane norm of the point $(8,0)$. Norm is a function $N:\mathbb{C}\to \mathbb{R}$, it takes value from complex plane and gives the euclidean distance from origin(real number) as output. Here, $|8|$ is $8$, as it is on the real line.

How can I rewrite the above set of equations so I don't have this kind of confusion.

You can consider $z=8^{1/3}$ and then you will get the equation $z^3-8=0$, or, $$(z-2)(z^2+2z+4)=0$$.

tarit goswami
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  • Thanks for the answer. But I wonder what I am doing wrong with my set of equations. Isn't $8 \times 1 = 8$? Isn't $1 = e^{i 2 \pi k}, k \in \mathbb{Z}$? Isn't then $8 = 8 \times e^{i 2 \pi k} = |8| \times e^{i 2 \pi k} $? Then I am just taking square root on both sides. – Olórin Sep 21 '18 at 18:53
  • They are not all same number, they are three different numbers, you can't write $8^{1/3}=...$, actually if $a_0,a_1,a_2$ are the cube roots of $8^{1/3}$ then, $$a_i=|8|^{1/3}\cdot e^{2\pi i/3}$$ with $i=0$ you only have $8^{1/3}=|8|^{1/3}$, as, it makes, $e^{2\pi i/3}=e^{2\pi\cdot 0 /3} = 1$ – tarit goswami Sep 21 '18 at 18:57