9
$\begingroup$

This may be a very simple question to ask, but I am confused with these definitions and would like to clarify here.

$\sqrt{81} = 9$. But $\sqrt{81} \ne -9$ because $\sqrt{}$ is used to represent the principal root.

So, If I want to represent both the roots, I have to mention it as $\pm\sqrt{81} = \pm 9$.

We know every real number ($\ne 0$) has three cube roots, one real and two complex. So, if we say $\root 3 \of {27}$, it means the principal cube root, which is $3$.

If so,

(a) How do we indicate that we are referring to all the three cube roots together (like $\pm \sqrt{81}$ for square roots) because $\root 3 \of {}$ refers to only principal cube root?

(b) Why are there no commonly accepted guidelines to decide the principal cube root because in some places, it is the real number while some books refer it to the one in positive imaginary axis.

Please help me to clear my doubts.

$\endgroup$
3
  • $\begingroup$ An alternative to using surds to write nth roots is to use Euler's formula instead e.g. $e^{i 2\pi/3}$ is the cubic root of unity with positive imaginary part. $\endgroup$ Feb 8, 2017 at 23:02
  • $\begingroup$ some books refer it to the one in positive imaginary axis Do you have an example of such a book? That would be an odd definition, since such a root needs not exist in general. $\endgroup$
    – dxiv
    Feb 8, 2017 at 23:07
  • $\begingroup$ I am also confused by this remark about "the one in positive imaginary axis." Notice for example that $(-i)^3 = i$. I tried the Wolfram Alpha query cube root of i and it identified a complex root as the principal root, when I thought it would identify the purely imaginary root as the principal root. What's your take? Which one do you feel should be the principal cubic root of $i$? $\endgroup$
    – Lisa
    Feb 17, 2017 at 22:26

3 Answers 3

10
+100
$\begingroup$

As you noticed yourself, we use $\pm\sqrt{x}$ to denote both square roots of some number $x$. Unfortunately there isn’t an analogous symbol to $\pm$ for cube roots, that allows you to write all three roots with a single symbol.

When we want to do this anyway, we usually write the following: \begin{align} y^3 &= x \\ y &= \sqrt[3]{x}\cdot\mathrm e^{2\pi in/3} & \text{where $n\in\mathbb Z$} \end{align} Now, remember that $\mathbb Z$ means the set of integers. If you compute the formula above when $n=0$, you’ll just get the principal root, since the exponent turns into $\mathrm e^0 = 1$.


If you instead compute it for $n=1$, you get something else: \begin{align} \sqrt[3]{x}\cdot\mathrm e^{2\pi i\cdot1/3}=\sqrt[3]{x}\cdot\mathrm e^{2\pi i/3} \end{align} If you happen to know a bit about the complex exponential function, you know that $\mathrm e^{2\pi i} = 1$, so if we cube the value above, we get: \begin{align} (\sqrt[3]{x}\cdot\mathrm e^{2\pi i/3})^3 = x\cdot\mathrm e^{2\pi i} = x\cdot 1 = x \end{align} So that was also a cube root of $x$.


What about $n=2$? \begin{align} \sqrt[3]{x}\cdot\mathrm e^{2\pi i\cdot2/3}=\sqrt[3]{x}\cdot\mathrm e^{4\pi i/3} \end{align} Using the same formula $\mathrm e^{2\pi i} = 1$ as last time, we get: \begin{align} (\sqrt[3]{x}\cdot\mathrm e^{4\pi i/3})^3 = x\cdot\mathrm e^{4\pi i} = x\cdot\mathrm (e^{2\pi i})^2 = x\cdot 1^2 = x \end{align} Yet another cube root of $x$!


What about other values of $n$? Since $n$ is an integer, we can write it as $n = 3a+b$ for some other integers $a$ and $b$, where $b$ is either $0$, $1$, or $2$. Now notice the following: \begin{align} \sqrt[3]{x}\cdot\mathrm e^{2\pi in/3} &= \sqrt[3]{x}\cdot\mathrm e^{2\pi i(3a+b)/3} \\ &= \sqrt[3]{x}\cdot\mathrm e^{2\pi i(3a)/3}\cdot\mathrm e^{2\pi ib/3} \\ &= \sqrt[3]{x}\cdot\mathrm e^{2\pi ia}\cdot\mathrm e^{2\pi ib/3} \\ &= \sqrt[3]{x}\cdot(\mathrm e^{2\pi i})^a\cdot\mathrm e^{2\pi ib/3} \\ &= \sqrt[3]{x}\cdot1^a\cdot\mathrm e^{2\pi ib/3} \\ &= \sqrt[3]{x}\cdot\mathrm e^{2\pi ib/3} \\ \end{align} Now since $b$ is either $0$, $1$ or $2$, any other value of $n$ is going to make the formula result in one of the three things we got earlier.


Are those three things from earlier actually different? It just so happens that we can write the following: \begin{align} \mathrm e^{2\pi i/3} &= -\frac12 + i\frac{\sqrt3}2\\ \mathrm e^{4\pi i/3} &= -\frac12 - i\frac{\sqrt3}2 \end{align} As we can see in the above formulas, they are in fact different. From this information we can conclude that the formula $\sqrt[3]{x}\cdot\mathrm e^{2\pi in/3}$ where $n\in\mathbb Z$ gives three different cube roots of $x$, and since we know there are exactly three, this is all of the cube roots.


Alternative notation and other roots

Some people define the number $\omega$ the following way: $$ \omega = -\frac12 + i\frac{\sqrt3}2 $$ Since we know $\mathrm e^{2\pi i/3} = -\frac12 + i\frac{\sqrt3}2$, this means that we can rewrite $\sqrt[3]{x}\cdot\mathrm e^{2\pi in/3}$ to the following: \begin{align} \sqrt[3]{x}\cdot\mathrm e^{2\pi in/3} = \sqrt[3]{x}\cdot\omega^n \end{align} Since it’s shorter, this notation appears better at a first glance, but there is actually a good reason to use the exponential function: it generalizes better. Take a look: \begin{align} y^{29} &= x\\ y &= \sqrt[29]{x}\cdot\mathrm e^{2\pi in/29} & \text{where $n\in\mathbb Z$} \end{align} Now, if we had settled on using $\omega$, we would have run into the same problem as the one we ran into when trying to use the cube roots, but with the exponential function, it works for any natural number.


Cube roots of complex numbers, and which root is principal?

When $x$ is a real number, it’s rather easy to pick a principal cube root: just pick the one on the real axis. Similarly for other roots, if the root number is odd, there is always exactly one on the real axis, and if the root number is even, there is always exactly one on the positive real axis.

But what if $x$ isn’t a real number? There are still 3 cube roots, but which one is principal? You can make up additional rules that work in some cases, for example if $x$ is purely imaginary, pick the one on the imaginary axis.

It turns out that when you’re working with roots of complex numbers, you always want to talk about all the roots, not some particular one of them. Luckily the formulas I gave in the start of this post work no matter what cube root you use in place of $\sqrt[3]{x}$, so normally you don’t actually define a “principal root” when working with complex numbers; you just say $\sqrt[3]{x}$ is one of them, and use the formula I gave in order to get all of them.


Additional comments

I’ve ignored zero in this post. In the case of zero, the formula just always gives $0$, no matter which $n$ you choose, so it still works as expected.

There is actually a way to unambiguously define the principal root of any complex number $x$. If you let $\arg(z)$ be the argument of the complex number $z$, then there is exactly one complex number $z$ such that $0\le\arg(z)<2\pi/3$ and $z^3 = x$. It works similarly for other roots of $x$.

$\endgroup$
1
  • $\begingroup$ Thanks. This cleared all my doubts. $\endgroup$
    – Kiran
    Feb 18, 2017 at 3:17
7
$\begingroup$

There seems to be some consensus among humans that $$\omega = \frac{-1}{2} + \frac{\sqrt{-3}}{2}$$ represents a complex cubic root of $1$. (Me personally I sometimes like to use 垰 for this purpose, just to be cryptic). Then you can use $\omega$ to help you represent the complex cubic roots of other real numbers. In a pinch, you can even use $w$, but then it would probably be better to write out "omega".

For example, $\root 3 \of 2$ represents the real cubic root of $2$, $(\root 3 \of 2) \omega$ represents the complex cubic root in the negative-positive quadrant of the complex plane, and $(\root 3 \of 2) \omega^2$ represents the complex cubic root in the negative-negative quadrant of the complex plane.

Or, to use your example of $27$, we have $3$ as the real cubic root, $3 \omega$ and $3 \omega^2$ are the complex cubic roots. Try this query in Wolfram Alpha: (3(-1/2 + sqrt(-3)/2)^2)^3, and then try it after deleting ^2.

$\endgroup$
2
$\begingroup$

When using the cubic formula (like the quadratic formula, except for cubics), the first root is found using the principal cube root, and in this case it is the one with the LARGES REAL PART, if it is a tie, you chose the one with the larges imaginary that is in the tie. The second root is the one with the largest imaginary part that is not the principle cube root. The third is the other. If you do not do this, you will not get the right answer in the cubic formula.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.