- #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.