What is wrong with the following proof

  • Thread starter snipez90
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  • #1
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What is wrong with the following "proof"

My friend was messing around and sent me something that leads to an incorrect conclusion:

e^iπ = -1
(e^iπ)^2 = (-1)^2
e^2iπ = 1
ln(e^2iπ) = ln1
2iπ * lne = ln1
2iπ * 1 = 0
2iπ = 0

After refreshing my memory with some of the most important theorems involving complex numbers, I'm still trying to find wrong step. My first qualm deals with lines 1-3. He puts his faith behind simple algebra, but I think it's just a coincidence. Line 1 and line 3 hold because of Euler's formula, which involves trig. Therefore, I believe that squaring to get 1 is a coincidence but then again according to the rules of algebra that should be valid?

Then the natural log part. I think that ln(e^2iπ) = ln[cos(2π) + i*sin(2π)] is undefined. But then my friend points out the obvious fact that the i*sin(2π) term vanishes, leaving ln(1), which of course is 0. I don't know too much about complex numbers, so any opinions on this discussion would be great.
 

Answers and Replies

  • #2
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While ln(e^2iπ) = ln1 is true, 2iπ = ln 1 is not. The natural logarithm of e^theta*i is defined as sigma*i where sigma is the corresponding angle lesser than 2pi. Therefore ln(e^2iπ) = 0*i and not 2iπ.
 
  • #3
mathman
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Strictly speaking the ln function is multivalued.

e2n(pi)i=1 for any n. Therefore ln(1)=2n(pi)i, for all n.
 
  • #4
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1) ln(e^2iπ) = ln1
2) 2iπ * lne = ln1
3) 2iπ * 1 = 0
The only problem is the top 2 lines. e^(ix) is not invertible. The single valued ln function is not its inverse.
 

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