How to find the cube roots of a real number not equal to 1?

[zenterix]

Homework Statement

If we want to find the cube roots of the number 1 we do
$$z^3=1=e^{2\pi n i}$$
$$z=e^{\frac{2\pi}{3}ni}$$
and for $n=0,1,2$ we obtain the three cubic roots of $1$.

How do we find the cube roots of, say, the number 5?

Relevant Equations

$$z^3=5=5e^{2\pi ni}$$

$$z=5^{1/3}e^{\frac{2\pi}{3}ni}$$

There is that term $5^{1/3}$ but that is exactly what we're trying to find.

 

[Vanadium 50]

Can you show your attempt?

 

[jedishrfu]

Of course, one of the cube roots of 5 is the cube root of 5 but there are two more that are 120 degrees apart on the complex plane if you were to plot them.

Has your teacher never shown you De Moivre's theorem?

 

[zenterix]

I think my confusion is arising because I am not taking into account that we are searching for the roots in the space of complex numbers.

When we write

$$x^3=5=5e^{2\pi ni}\tag{1}$$

$$x=5^{1/3}e^{\frac{2\pi}{3}ni}\tag{2}$$

the number $5^{1/3}$ in the expression above is a real number.

It is also a cube root of $5$.

We are searching for $5^{1/3}$ but in the complex space. For a real root, the root also has the representation $5^{1/3}$.

In other words, and please correct me if I am wrong, but we are using $5^{1/3}$ to denote two distinct things: (1) any complex root of $5$ and (2) the real root of $5$.

To find all the roots, all we have to do is use $n=0,1,2$ in (2) to find that

- for $n=0$ we get the real root $5^{1/3}$

- for $n=1$ we get the complex root $5^{1/3}e^{\frac{2\pi}{3}i}$

- for $n=2$ we get the complex root $5^{1/3}e^{\frac{4\pi}{3}i}$.

 

[Tom.G]

https://www.google.com/search?hl=en&q=hand+calculation+of+cube+root

(by the way, it took several weeks of asking my high school Algebra teacher how that is done. Of course I have long forgotten how, These days I just pick up a pocket calculator!)

Cheers,
Tom

 

[jedishrfu]

Visually what is the n=0,1,2 doing for

$$x=5^{1/3}e^{\frac{2\pi}{3}ni}\tag{2}$$

in the complex plane?

What does ${2\pi}/3$ equate to in degrees?

 

[zenterix]

Visually what is the n=0,1,2 doing for

$$x=5^{1/3}e^{\frac{2\pi}{3}ni}\tag{2}$$

in the complex plane?

These are also the three cube roots of the number 1. We multiply by $5^{1/3}$ to get the cube roots of $5$.

 

[zenterix]

My next question was going to be about something I read here.

Namely, that to solve $z^n=c=re^{i\theta}$ we can find one solution $r^{1/n}e^{i\theta/n}$ and then find all solutions by multiplying by the solutions to $z^n=1$.

The link seems to imply we can choose any one solution to then multiply by the solutions to the n-th roots of $1$.

Indeed, after checking this method for $z^3=5$ it does indeed work.

So basically, it seems that to solve $z^n=c$ we can take the real solution $c^{1/n}$ and multiply it by the $n$ $n$-th roots of unity.

 

[gmax137]

https://www.google.com/search?hl=en&q=hand+calculation+of+cube+root

(by the way, it took several weeks of asking my high school Algebra teacher how that is done. Of course I have long forgotten how, These days I just pick up a pocket calculator!)

Cheers,
Tom

We did it with log tables. 10^log(5)/3

 

[jedishrfu]

or with loglog decitrig bamboo sliderules.