Finding Arbitrary nth Roots of -1 Algebraically

[soandos]

is there a way to find all of the arbitrary nth roots of -1 algebraically?

 

[adriank]

Assuming you want to find roots in \mathbb{C}, if x^n = -1, then \lvert x \rvert = 1, so x = e^{i \theta} for some \theta. Since -1 = e^{i \pi}, all you need to do is find n distinct solutions of e^{i \theta} = e^{i \pi}. (That you only need n solutions follows from that you're effectively asking for the roots of the polynomial x^n + 1.)

 

[soandos]

how would that get me say, 1/2 (1 - I Sqrt[3])?

 

[Hurkyl]

is there a way to find all of the arbitrary nth roots of -1 algebraically?

You need to be more clear about what you mean by "find". e.g. why is (1/2) (1 - i 3^{1/2}) a pleasing answer, but not (\sqrt[3]{-1})^5?

 

[robert Ihnot]

Well, if you want to find them in terms of signs and cosins, we just use de Moivre's Theorem.

Since cos(pi) + isin(pi) = -1. We find the nth root as cos(pi/n) + isin(pi/n) = -1^1/n.
Other roots are powers of this root and we obtain 1/n, 2/n, 3/n...n/n=1. For n=3, the first root we encounter is cos(60) + isin(60)=1/2 + i(sqrt3)/2.

 

[MathematicalPhysicist]

Yes, as someone noted already, -1=exp(i*pi) take the nth root, i.e:
exp(i*pi+2pi*k)/n=(-1)^1/n.
I am being succint today, aren't I.
(-:

 

[HallsofIvy]

is there a way to find all of the arbitrary nth roots of -1 algebraically?

What do you mean by "algebraically"? By DeMoivre's theorem, we know that some roots necessarily involve complex exponentials or, equivalently, sine and cosine.

Generally, the nth roots of -1 are given by
cos\left(\frac{(1+ 2k)\pi}{n}\right)+ i sin\left(\frac{(1+ 2k)\pi}{n}\right)
where k ranges from 1 to n to give all n roots.

For example, since -1= e^{\pi}= cos(\pi)+ i sin(\pi) or could be written as -1= e^{\pi+ 2\pi}= e^{3\pi}= cos(3\pi)+ i sin(3\pi) or -1= e^{\pi+ 2\pi+ \2pi}= e^{5\pi}= cos(5\pi)+ i sin(\5pi), its cube roots are
e^{pi/3}= cos(\pi/3)+ i sin(\pi/3),
which equals 1/2+ i\sqrt{3}/2

e^{3pi/3}= cos(\pi)+ i sin(\pi)= -1, and

e^{5\pi/3}= cos(5\pi/3)+ i sin(5\pi/3),
which equals 1/2- i\sqrt{3}/2

Since \pi/3= 60 degrees, you can, of course, get those values by bisecting an equilateral triangle.

Similarly, you could get those values geometrically by recognizing that the roots lie on the vertices of an equilateral triangle inscribed in the unit circle.

But roots like the 5th root or 7th root will not give such nice values. If you do not include trig functions in "algebraically", there is no way.