My Math Forum Help with Euler's identity; i = 0??

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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, 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. Thanks from Tau
 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
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Quote:
 Originally Posted by skipjack What does "a rotation that settles to 1" mean?
The visual interpretation of the identity on the complex plane.

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