Solution to x^n = 1?

Question

Suppose that x^2 = 1. Then it follows that the real solutions of x are x = {1, -1}.

Now, suppose that x^3 = 1. In this case, x has only one real solution: x = 1.

For x^4 = 1, x has two solutions once more: x = {1, -1}.

It appears that for odd integers n, x^n = 1 has one real solution.
For even integers n, x^n = 1 has TWO real solutions.

My question is regarding values of n that do not fall under that category. Particularly, if n isn't an integer, yielding two cases:

1) What if n is a non-integer fraction? Will x have one real solution, two real solutions, or neither?

2) What if n is an irrational number (such as e, pi)? How many real solutions (if any) will x^n = 1 have?

I ask this to satisfy my mathematical curiosity, and if anybody can help me by answering those questions I would be grateful. Please give me as much details a possible when answering, as I'm interested in the mathematics behind it.

Note that I'm talking only about the REAL solutions.

Thanks in advance.

Answer 1

Interesting question.

After thinking about it, I believe that you ONLY get the two real roots {-1, 1} when n is an EVEN INTEGER (not equal to zero). Anything else you only get 1.

I base this on DeMoivre's formula (http://en.wikipedia.org/wiki/De_Moivre's_formula ) as follows.

x^n = 1

is equivalent to:

x = 1^(1/n)

Since 1 is written in complex form as 1(cos 0 + i sin 0), DeMoivre's formula says that the roots would be:

1^(1/n) = cos(2kπ/n) + i sin(2kπ/n)

Regardless of the value of n, you'd always have the default solution of 1 when k = 0...and then lots of other complex solutions.

But the ONLY times when you would have BOTH solutions of 1 and -1 are when n is an integer multiple of 2 other than zero (i.e. even). When that happens, it's guaranteed that for some k you will end up with cos π + i sin π = -1.

Proof:

You can only get -1 when the stuff inside the parentheses (2kπ/n) is equal to π. (And n is not equal to zero, of course.)

2kπ/n = π
2kπ = nπ
2k = n

Since k is "any integer", n must be "any even integer" (except zero, for which 2kπ/n is infinite). QED.

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Alnitaka's assertion that you'd have both roots for any p/q where q is even is wrong.

Easiest case: p/q = 1/2.

1^(1/2) = 1....but (-1)^(1/2) does not equal 1! It's i.

Again looking at DeMoivre's formula:

cos(2kπ/n) + i sin(2kπ/n)

Plugging in n = 1/2:

cos(4kπ) + i sin(4kπ)

Regardless of what value of k you choose, you can't get the odd multiple of π that would you need in order to get cos π = -1.

Answer 2

You want only real solutions, but it helps to consider all solutions, including complex ones, because an interesting pattern emerges.

x^3 = 1 is a cubic equation so it has three roots, namely 1 and (-1 +- i sqrt(3))/2. If you graph these on the complex plane, they form an equilateral triangle.

x^5 = 1 has five roots, namely 1 and 4 complex numbers expressible in terms of square roots, sqrt(5) and i. These form a regular pentagon on the complex plane.

One way of looking at this is to observe that all roots of x^n = 1 will lie on the unit circle; that is, they will be a+ib, where a^2+b^2 = 1. If it is cube roots, you go around a third of the circle each time. This gives you the triangle. With fifth roots, you go a fifth circle each time, producing the pentagon. With square roots, you go half the circle, giving two opposing points - 1 and -1.

So we can answer your questions now. If n = p/q, then each time around, we will go a qth of the circle around p times. It does not matter whether you go once or p times, going around a qth of the circle each time will keep you on the same q-gon. If q is prime, then once again you will describe a q-gon, and so the only real root is 1; the others are off the real axis. Assume p/q is in lowest terms. If q is odd, you will not go on the real line except for the starting point. Hence the only real solution to x^n =1 is 1. If q is even, then by taking a suitable number of times around you get halfway around and land at -1. So the solutions in this case are 1 and -1.

2. If n is irrational, you will never go exactly halfway around the circle, since that implies that n is rational. So you will never hit the -1 point. So in this case the only real root is 1. There are a countable infinity of other solutions, all of them complex.

To sum up: if n = p/q in lowest terms with q even, the solutions are -1 and 1. Otherwise, the equation has only 1 for a real root.

Answer 3

If you want only real solutions, you're stuck with what you have listed, with multiple roots.
Just as the equation (x^2 - 4x + 4) = (x - 2)^2 has a solution x = 2 of multiplicity 2, so also the equation x^n has 2 solutions of multiplicity n/2 for even values of n.
This doesn't seem like a satisfactory answer, and it wasn't, to mathematicians such as C.F. Gauss. This was part of the motovation for inventing complex numbers. If you allow complex numbers you'll find the expected number of roots for equations like yours. Moreover, if you graph them, you'll find them nicely distributed, evenly around the unit circle.
These solutions are not "un-real" in the sense that you suggest. If you read Pi R Squared's explantion of complex numbers in another recnetly answered question, you'll find that complex numbers can be quite real (in the usual sense of the word, not the mathematical sense.)

Answer 4

x=1, n=1

Answer 5

1 =1e^(0 + 2m π i), where m is any integer.

1^(1/n) = 1e^[2(m/n)π i], where n is any real number.

Therefore, whenever m/n = 0, or multiple of 1/2, you get real solution. Otherwise, no real solution can be found.

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It seems to me all the discussions below is somewhat the same as I discussed above. It is not that complicated!