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 May 29th, 2019, 09:17 PM #1 Member   Joined: Aug 2018 From: România Posts: 88 Thanks: 6 Comparison Hello all, Which number is greater , $\displaystyle (-1)^i$ or $\displaystyle (-1)^{-i}$ where $\displaystyle i^2=-1$? All the best, Integrator Last edited by Integrator; May 29th, 2019 at 09:20 PM.
 May 29th, 2019, 09:46 PM #2 Member     Joined: Oct 2018 From: USA Posts: 90 Thanks: 61 Math Focus: Algebraic Geometry $a^{x} = (e^{\ln{a}})^{x} = e^{x \ln{a}}$ $(-1)^{i} = e^{i \ln{(-1)}}$ Using the identity $e^{i \pi} = -1$ , $i \pi = \ln{(-1)}$ $e^{i \ln{(-1)}} = e^{i^{2} \pi} = \frac{1}{e^{\pi}}$ $(-1)^{-i} = \frac{1}{(-1)^{i}} = e^{\pi}$ Since $e^{\pi} > \frac{1}{e^{\pi}}$ , $(-1)^{-i} > (-1)^{i}$ Last edited by Greens; May 29th, 2019 at 09:48 PM. Reason: Formatting
 May 29th, 2019, 09:46 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 642 Thanks: 406 Math Focus: Dynamical systems, analytic function theory, numerics The complex numbers are not totally ordered. There is no reasonable notion of "greater than" so the question is meaningless. Thanks from topsquark
May 30th, 2019, 03:07 AM   #4
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Quote:
 Originally Posted by Greens $a^{x} = (e^{\ln{a}})^{x} = e^{x \ln{a}}$ $(-1)^{i} = e^{i \ln{(-1)}}$ Using the identity $e^{i \pi} = -1$ , $i \pi = \ln{(-1)}$ $e^{i \ln{(-1)}} = e^{i^{2} \pi} = \frac{1}{e^{\pi}}$ $(-1)^{-i} = \frac{1}{(-1)^{i}} = e^{\pi}$ Since $e^{\pi} > \frac{1}{e^{\pi}}$ , $(-1)^{-i} > (-1)^{i}$
The reason this didn't work right is that the exponential function in the complex plane takes on multiple values. The best way to represent this is to say that the exponential function "spirals". There are several copies, called "sheets". For example, there is a sheet with $\displaystyle 0 \leq \theta < 2 \pi$, another with $\displaystyle 2 \pi \leq \theta < 4 \pi$, etc. These sheets are identical and you can't say that one is greater than the other because it is on a "higher" or "lower" sheet.

Calculationally speaking, $\displaystyle e^{i \pi} = -1$ but so does $\displaystyle e^{i \pi + 2n \pi} = -1$ and thus $\displaystyle \ln(-1) = i \pi + 2n \pi$ for any values of n you like.

-Dan

Addendum: I messed with this imaginary exponent thing a while back and it just didn't work well. I won't say you can't do anything with complex powers but it is definitely very messy.

Last edited by skipjack; May 30th, 2019 at 04:29 PM.

May 30th, 2019, 06:10 AM   #5
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Quote:
 Originally Posted by topsquark messy.
I agree completely. As I googled my way through problem I realized there was probably way more going on than I knew about, but I figured it wouldnt hurt to post something and figure it out.

Begs the question, where did this problem come from?

 May 30th, 2019, 01:28 PM #6 Member     Joined: Oct 2018 From: USA Posts: 90 Thanks: 61 Math Focus: Algebraic Geometry So I came back to this and now I'm curious. Using $\ln{(-1)} = i\pi + 2n\pi$ $(-1)^{i} = e^{i \ln{(-1)}} = e^{i(i\pi + 2n\pi)} = e^{\pi(2n-1)}$ Then $(-1)^{-i} = e^{-\pi(2n-1)}$ Solve for $n$ equality: $-\pi(2n-1) = \pi(2n-1)$ $n= \frac{1}{2}$ So $e^{\pi(2n-1)} > e^{-\pi(2n-1)}$ for $n > \frac{1}{2}$ , but since $n$ shouldn't be a fraction we can just say $n \geq 1$ So this would mean $(-1)^{i} > (-1)^{-i}$ for sheets such that $n \geq 1$ and $(-1)^{-i} > (-1)^{i}$ on the $n=0$ sheet. Is this the correct way to go about this or is there something I'm missing? Last edited by skipjack; May 30th, 2019 at 01:47 PM.
May 30th, 2019, 02:08 PM   #7
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Quote:
 Originally Posted by Greens Using $\ln{(-1)} = i\pi + 2n\pi$
Did you mean $\ln(-1) = i(\pi + 2n\pi)$?

May 30th, 2019, 02:44 PM   #8
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 Originally Posted by topsquark $\displaystyle ln(-1) = i \pi + 2n \pi$ for any values of n you like.
This is the value I was using. I do believe you're right, skipjack, I'll go work it with this value and see what comes up.

EDIT:

$(-1)^{i} = e^{i \ln{(-1)}} = e^{i^{2} (\pi + 2n \pi )} = e^{- \pi (2n+1)}$

$(-1)^{-i} = e^{ \pi (2n+1)}$

$2n+1 = -2n-1$

$n = -\frac{1}{2}$ So we can just say $e^{ \pi (2n+1)} > e^{- \pi (2n+1)}$ for $n \geq 0$. Therefore $(-1)^{-i} > (-1)^{i}$

I believe I've developed an $i$-phobia.

Last edited by skipjack; May 30th, 2019 at 04:07 PM. Reason: Clarity

May 30th, 2019, 03:39 PM   #9
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Quote:
 Originally Posted by skipjack Did you mean $\ln(-1) = i(\pi + 2n\pi)$?
That was my bad. It's a typo from my post.

-Dan

May 30th, 2019, 04:26 PM   #10
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Quote:
 Originally Posted by Greens $2n+1 = -2n-1$
Why do that? As you showed, it has no integer solution.

If you change $n$ to $-n - 1$, $2n+1$ becomes $-2n - 1$. They both represent an arbitrary odd integer.

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