MHB Why are the cubic roots of 1 expressed as cis 120 degrees?

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The cubic roots of 1 include not only the real root 1 but also two complex roots, which can be expressed as cis(120°) and cis(240°). These complex roots arise from solving the equation x^3 = 1, leading to the factorization x^3 - 1 = (x - 1)(x^2 + x + 1). The quadratic equation x^2 + x + 1 yields complex solutions that correspond to angles of 120° and 240° on the unit circle. This illustrates that the n-th roots of unity are evenly spaced around the unit circle, highlighting the connection between complex numbers and geometric representations. Understanding these roots deepens the appreciation of the relationship between algebra and geometry in mathematics.
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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 ?
 
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Let's let:

$$x^3=1=e^{2\pi ki}$$

Hence:

$$x=e^{\frac{2}{3}\pi ki}=\text{cis}\left(\frac{2k}{3}\pi\right)$$
 
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 ?
 
Yankel said:
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:

$$x^4=1$$

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 $$\theta=\frac{2k\pi}{n}$$ with $0\le k<n,\,k\in\mathbb{Z}$.
 
Yankel said:
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.
 
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