Complex Analysis Complex Analysis Math Forum

 May 29th, 2019, 10:17 PM #1 Senior Member   Joined: Aug 2018 From: România Posts: 112 Thanks: 7 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 10:20 PM. May 29th, 2019, 10:46 PM #2 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 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 10:48 PM. Reason: Formatting May 29th, 2019, 10:46 PM #3 Senior Member   Joined: Sep 2016 From: USA Posts: 684 Thanks: 459 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, 04:07 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 05:29 PM. May 30th, 2019, 07:10 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, 02:28 PM #6 Senior Member   Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 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 02:47 PM. May 30th, 2019, 03:08 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, 03:44 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 05:07 PM. Reason: Clarity May 30th, 2019, 04:39 PM   #9
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 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, 05:26 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. Tags comparison Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Alexis87 Algebra 1 July 31st, 2013 01:38 PM zaff9 Calculus 1 April 18th, 2013 07:05 AM vevangelist Economics 2 May 22nd, 2012 07:24 AM fuzzwaz Calculus 3 November 5th, 2011 03:34 PM md_2674062 Calculus 6 July 20th, 2011 12:45 PM

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