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May 25th, 2015, 06:49 PM  #1 
Member Joined: Jan 2014 Posts: 86 Thanks: 4  Help with Euler's identity; i = 0??
 e^(i2π) = 1  e^(i2π) = e^0  i2π = 0  i = 0 (?) Where is the mistake? 
May 25th, 2015, 07:38 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
You are equating a principal value with a nonprincipal value. $$\newcommand{\e}{\mathrm e} \newcommand{\i}{\mathrm i} \e^{2\pi\i} = 1 = \e^{0 + 2n\pi\i} \\ 2\pi\i = 2n\pi\i \\ n = 1 $$ Also, the complex logarithm is multivalued and $\log 1 = 2n\pi\i$. 
May 25th, 2015, 08:01 PM  #3 
Member Joined: Jan 2014 Posts: 86 Thanks: 4 
I get it. So is it correct to also distinguish them qualitatively? To say that with the complex exponent we have a rotation that settles to 1 while for n=0 we don't have a rotation to settle to 1?
Last edited by skipjack; May 25th, 2015 at 09:42 PM. 
May 25th, 2015, 08:11 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
I think so, yes. It is rare that we are interested in any but the principal values anyway.

May 25th, 2015, 09:43 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 21,128 Thanks: 2337 
What does "a rotation that settles to 1" mean?

May 26th, 2015, 04:34 AM  #6 
Member Joined: Jan 2014 Posts: 86 Thanks: 4  

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