6 ms·
Help! What do I do wrong: e^(πi) = -1 <=> e^(2πi) = 1 <=> ln(e^(2πi)) = ln(1) <=> 2πi = 0 <=> i = 0
by jonsen 16y ago
Help! What do I do wrong:
e^(πi) = -1 <=>
e^(2πi) = 1 <=>
ln(e^(2πi)) = ln(1) <=>
2πi = 0 <=>
i = 0
- skybrian 16y agoLogarithms are tricky with complex numbers because rotation wraps around if you go 360 degrees (2 pi i).
- scotty79 16y agoln(e^(2πi)) is not 2πi http://www.wolframalpha.com/input/?i=ln(e^(2%CF%80i) http://www.wolframalpha.com/input/?i=ln(e^(2%CF%80i)) It's similar mistake to claiming that sqrt(square(-2)) is -2
- lotharbot 16y agoThe complex log function is a "branching function" [1] with an infinite number of branches. For comparison, the square root function has two branches. sqrt(x^2) has two values, +x and -x. ln(e^x) has an infinite number of values, x + 2nπi (for any integer n). 2πi = 0 + 2nπi for some n. (You can repeat this for, say, e^4πi, e^989781123972πi, e^-091784πi, etc. Those are all equal to 1.) [1] http://en.wikipedia.org/wiki/Complex_logarithm http://en.wikipedia.org/wiki/Complex_logarithm