Understanding Cubic Roots of 1: Exploring cis 120k

[Yankel]

Hello

I didn't know in which forum to put this...

I solved a linear algebra question, and my answer was:

{1}^{1/3}

which to my understanding is 1. In the book however, they said it is equal to cis 120k k=0,1,2,...

where 120 is degrees. I tried taking the complex number 1+0i and turn it into it's polar version but did not get 120 degrees. Can you explain to me why the cubic root of 1 is cis 120 ?

 

[MarkFL]

Let's let:

\(\displaystyle x^3=1=e^{2\pi ki}\)

Hence:

\(\displaystyle x=e^{\frac{2}{3}\pi ki}=\text{cis}\left(\frac{2k}{3}\pi\right)\)

 

[Yankel]

I see...

but why did they do it in the first place ?

Am I wrong that the cubic root of 1 is 1 ?

What about the quarter root of 1 ?

 

[MarkFL]

I see...

but why did they do it in the first place ?

Am I wrong that the cubic root of 1 is 1 ?

What about the quarter root of 1 ?

If you restrict yourself to real roots, then $x=1$ is the only such real root. However, as I am sure you know, a cubic equation will have 3 roots, and since there is only 1 real root to the equation in question, we know there must be two complex roots, and we know further that they are conjugates.

The quarter or 4th root of 1 will satisfy:

\(\displaystyle x^4=1\)

You can solve this by factoring. In general we will find the $n$th roots of unity to be equally spaced about the unit circle in an Argand diagram, where \(\displaystyle \theta=\frac{2k\pi}{n}\) with $0\le k<n,\,k\in\mathbb{Z}$.

 

[Deveno]

I see...

but why did they do it in the first place ?

Am I wrong that the cubic root of 1 is 1 ?

What about the quarter root of 1 ?

You're not wrong, but you're not entirely right, either.

First, what do we MEAN by: "a cube root of 1"?

We mean some "number" $x$ such that: $x^3 = 1$. Clearly, 1 works, since: $1^3 = 1$.

Another way to phrase this is:

$x^3 - 1 = 0$

Now, $x^3 - 1 = (x - 1)(x^2 + x + 1)$.

Taking $x = 1$, we see the left factor is 0, so the whole product is 0.

But what if $x \neq 1$? Could it be possible that $x^2 + x + 1 = 0$?

Well, if we use the quadratic formula, with $a = b = c = 1$, we obtain:

$x = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(1)}}{2} = \dfrac{-1 \pm \sqrt{-3}}{2}$

which can be written in the form:

$x = -\dfrac{1}{2} \pm i\dfrac{\sqrt{3}}{2}$.

Note that this is:

$x = \cos\left(\dfrac{2\pi}{3}\right) \pm i\sin\left(\dfrac{2\pi}{3}\right)$

in other words, in the complex plane the "other two cube roots of 1" lie at the angles:

1/3 around the circle, and 2/3 (-1/3) around the unit circle.

You can verify, by direct computation, that if:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$

that $\omega^2 = \overline{\omega}$, and also that:

$x^2 + x + 1 = (x - \omega)(x - \omega^2)$

(remember, complex solutions to a REAL quadratic come in conjugate-pairs).

The situation is quite analogous for the polynomial:

$x^n - 1$

the roots are:

$\cos\left(\dfrac{2k\pi}{n}\right) + i\sin\left(\dfrac{2k\pi}{n}\right)$

for $k = 0,1,2,\dots,n$.

For $n = 4$ (the fourth roots of 1), we get:

for $k = 0,\ \cos(0) + i\sin(0) = 1 + i0 = 1$.

for $k = 1,\ \cos\left(\dfrac{\pi}{2}\right) + i\sin\left(\dfrac{\pi}{2}\right) = 0 + i1 = i$

for $k = 2,\ \cos(\pi) + i\sin(\pi) = -1 + i0 = -1$

for $k = 3.\ \cos\left(\dfrac{3\pi}{2}\right) + i\sin\left(\dfrac{3\pi}{2}\right) = 0 +i(-1) = -i$.

Indeed, we have: $x^4 + 1 = (x^2 + 1)(x^2 - 1) = (x + i)(x - i)(x + 1)(x - 1)$.

It turns out there is a DEEP connection between $n$-th roots of a number, and $\frac{1}{n}$-th of a circle. The geometrical reason for this is that complex multiplication is "part stretching" and "part rotating".

The circle is a profound mathematical object. I cannot stress this enough. In mathematics we have two "big ideas": the line, and the circle. The extrapolation of these two simple things, leads to a vast array of interesting structures.