## Pi root of -1

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- Pied typer
**Posts:**22**Joined:**Fri Mar 11, 2016 1:59 am UTC**Location:**[i]somewhere[/i]

### Pi root of -1

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?

### Re: Pi root of -1

The definition of complex exponentiation is that a

Thus, the π’th root of −1 is found to be:

(−1)

^{b}= e^{b·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}= e^{Log(−1)/π}= e^{(iπ)/π}= e^{i}= cos(1) + i·sin(1)wee free kings

- Pied typer
**Posts:**22**Joined:**Fri Mar 11, 2016 1:59 am UTC**Location:**[i]somewhere[/i]

### Re: Pi root of -1

Qaanol wrote:The definition of complex exponentiation is that a^{b}= e^{b·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}= e^{Log(−1)/π}= e^{(iπ)/π}= e^{i}= 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 1

^{x}= 1, how is it that -1

^{1/π}= sin(1)i? That's even weirder.

Hey, is my flair

Also, go here:

http://stackoverflow.com/q/11227809/6388243

^{TM}small enough for here?Also, go here:

http://stackoverflow.com/q/11227809/6388243

### Re: Pi root of -1

What does 1

You asked about a power of (-1).

^{x}has to do with anything?You asked about a power of (-1).

- Pied typer
**Posts:**22**Joined:**Fri Mar 11, 2016 1:59 am UTC**Location:**[i]somewhere[/i]

### Re: Pi root of -1

PsiCubed wrote:What does 1^{x}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 1

^{1/π}can be that.

Hey, is my flair

Also, go here:

http://stackoverflow.com/q/11227809/6388243

^{TM}small enough for here?Also, go here:

http://stackoverflow.com/q/11227809/6388243

### Re: Pi root of -1

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. 1

^{1/π}is not the same as -1

^{1/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.

### Re: Pi root of -1

(-1)

^{x}= e^{x*Log(-1)}= e^{iπx}= 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.).she/they

gmalivuk wrote:Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.

- Pied typer
**Posts:**22**Joined:**Fri Mar 11, 2016 1:59 am UTC**Location:**[i]somewhere[/i]

### Re: Pi root of -1

Sizik wrote:(-1)^{x}= e^{x*Log(-1)}= e^{iπx}= 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.)

Hey, is my flair

Also, go here:

http://stackoverflow.com/q/11227809/6388243

^{TM}small enough for here?Also, go here:

http://stackoverflow.com/q/11227809/6388243

### Re: Pi root of -1

Also note that any complex z for which z

Thus if m is any odd integer, then e

^{π}= −1 can be considered a π’th root of −1.Thus if m is any odd integer, then e

^{m·i}= cos(m) + i·sin(m) is a π’th root of −1.wee free kings

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