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.

May 30th, 2019, 04:07 AM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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 
Senior Member Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry  
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 $\ln{(1)} = i\pi + 2n\pi$ $(1)^{i} = e^{i \ln{(1)}} = e^{i(i\pi + 2n\pi)} = e^{\pi(2n1)}$ Then $(1)^{i} = e^{\pi(2n1)}$ Solve for $n$ equality: $\pi(2n1) = \pi(2n1)$ $n= \frac{1}{2}$ So $e^{\pi(2n1)} > e^{\pi(2n1)}$ 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 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332  
May 30th, 2019, 03:44 PM  #8  
Senior Member Joined: Oct 2018 From: USA Posts: 102 Thanks: 77 Math Focus: Algebraic Geometry  Quote:
EDIT: $(1)^{i} = e^{i \ln{(1)}} = e^{i^{2} (\pi + 2n \pi )} = e^{ \pi (2n+1)}$ $(1)^{i} = e^{ \pi (2n+1)}$ $2n+1 = 2n1$ $n = \frac{1}{2}$ 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 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 986 Math Focus: Wibbly wobbly timeywimey stuff.  
May 30th, 2019, 05:26 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 21,124 Thanks: 2332  

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