Pi root of -1

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Pied typer
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Pi root of -1

Postby Pied typer » Fri Mar 11, 2016 2:07 am UTC

I'm wondering, since the existence of a real square root of a negative number depends on the numerator of the reduced fraction of the root, what happens if the root is irrational? So, in other words, is the pi-th root of -1 simplified to -1 or i?

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Qaanol
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Re: Pi root of -1

Postby Qaanol » Fri Mar 11, 2016 4:45 am UTC

The definition of complex exponentiation is that ab = eb·Log(a), where ‘Log’ is the principle branch of the natural logarithm (meaning for all z, Log(z) = u + i·v with −π < v ≤ π).

Thus, the π’th root of −1 is found to be:
(−1)1/π = eLog(−1)/π = e(iπ)/π = ei = cos(1) + i·sin(1)
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Re: Pi root of -1

Postby Pied typer » Fri Mar 11, 2016 1:59 pm UTC

Qaanol wrote:The definition of complex exponentiation is that ab = eb·Log(a), where ‘Log’ is the principle branch of the natural logarithm (meaning for all z, Log(z) = u + i·v with −π < v ≤ π).

Thus, the π’th root of −1 is found to be:
(−1)1/π = eLog(−1)/π = e(iπ)/π = ei = cos(1) + i·sin(1)

:| I had to look that over a couple times before I understood it. It's still difficult to do that nonetheless.

Wait... If 1x = 1, how is it that -11/π = sin(1)i? That's even weirder.
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Re: Pi root of -1

Postby PsiCubed » Fri Mar 11, 2016 2:33 pm UTC

What does 1x has to do with anything?

You asked about a power of (-1).
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Re: Pi root of -1

Postby Pied typer » Fri Mar 11, 2016 3:24 pm UTC

PsiCubed wrote:What does 1x has to do with anything?

You asked about a power of (-1).

(-1)2 = 1
This means that (-1)x can only ever evaluate to three results: 1, -1, or +-i. sin(1)i is not one of those; I'm wondering how 11/π can be that.
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Re: Pi root of -1

Postby jaap » Fri Mar 11, 2016 3:43 pm UTC

Pied typer wrote:(-1)2 = 1
This means that (-1)x can only ever evaluate to three results: 1, -1, or +-i.

No it does not mean that.
For example, consider the complex number (1+i)/sqrt(2). If you square this, you get:
[(1+i)/sqrt(2)]2 = (1+i)2/2 = (1*1 + 1*i + i*1 + i*i)/2 = (1 + 2i + -1)/2 = i
So this number to the fourth power is -1:
[(1+i)/sqrt(2)]4 = -1
So that complex number is one of the fourth-roots of -1, which you could write as:
(1+i)/sqrt(2) = (-1)1/4
So here is a power of -1 that is obviously not 1, -1, i or -i.

Note: Be careful with your minus signs. 11/π is not the same as -11/π, which is not the same as (-1)1/π.
Last edited by jaap on Fri Mar 11, 2016 3:46 pm UTC, edited 1 time in total.

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Re: Pi root of -1

Postby Sizik » Fri Mar 11, 2016 3:45 pm UTC

(-1)x = ex*Log(-1) = eiπx = the point in the complex plane that is distance 1 from the origin, and πx radians from the positive real axis. This is +1 for even integers, -1 for odd integers, +i for x = (4n+1)/2 (i.e. 1/2, 5/2, etc.), and -i for x = (4n-1)/2 (i.e. -1/2, 3/2, etc.).
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Re: Pi root of -1

Postby Pied typer » Fri Mar 11, 2016 3:58 pm UTC

Sizik wrote:(-1)x = ex*Log(-1) = eiπx = the point in the complex plane that is distance 1 from the origin, and πx radians from the positive real axis. This is +1 for even integers, -1 for odd integers, +i for x = (4n+1)/2 (i.e. 1/2, 5/2, etc.), and -i for x = (4n-1)/2 (i.e. -1/2, 3/2, etc.).

Huh. That's actually a pretty good way to understand it. Now I just draw a unit circle in the complex plane, and get what was said before. (And, actually, now that I notice it, I missed the cos(1) term in that.)
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Re: Pi root of -1

Postby Qaanol » Sat Mar 12, 2016 3:54 pm UTC

Also note that any complex z for which zπ = −1 can be considered a π’th root of −1.

Thus if m is any odd integer, then em·i = cos(m) + i·sin(m) is a π’th root of −1.
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