- #1
snipez90
- 1,101
- 5
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.
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.