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Using the relation $\ e^{ix} = \cos(x) + i\sin(x)$ and substituting for $\ x = \pi$, we have the well-known Euler identity, $ e^{i\pi} = -1$. Substitute also for $ x = -\pi $, we have $ e^{-i\pi} = -1$. So we can say $ e^{i\pi} = e^{-i\pi}$ and taking $\ln$ on both sides, $i\pi=-i\pi$. There is clearly something wrong here. Can some help me to figure out what's wrong?

EDIT : I actually had a confusing output from my calculator while trying to solve the problem : enter image description here

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The logarithm is not a function for complex numbers (as you discovered ;-) If you look at the values for $e^{i x}$, you see that it is periodic with period $2 \pi$, so the inverse isn't well defined. –  vonbrand Apr 19 '13 at 16:34
    
What do you mean saying The logarithm is not a function for complex numbers? My calculator gives me $ln(e^{x*i}) = x*i$ –  moray95 Apr 19 '13 at 16:48
    
Then your calculator is doing it wrong. What calculator is this? –  J. M. Apr 19 '13 at 16:49
    
It's TI NSpire CX Cas, the best one as I know. –  moray95 Apr 19 '13 at 16:50

3 Answers 3

up vote 2 down vote accepted

$e^{ix}=e^{iy}\implies e^{i(x-y)}=1=e^{2n\pi i}$ as $e^{2n\pi i}=\cos2n\pi+i\sin2n\pi=1$

$\implies x-y=2n\pi$ where $n$ is any integer

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Got your logic but still, can't see why mine is wrong... –  moray95 Apr 19 '13 at 16:44
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Your calculation can be true if and only if $ln$ be a one-to-one function and $ln$ is not. –  Hoseyn Heydari Apr 19 '13 at 17:02

The logarithm is multi-valued. So we define a branch cut of the logarithm as a principal branch along an axis (or any line) so that we only use one cycle of phase of period $2 \pi$; this keeps that principal branch as a single-valued function.

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In definition of $ln(z)$ for $z\in \mathbb{C}$ we have: $$ln(x+iy)=ln(|x|)+iArg(y)$$ Since $Arg$ can achive multi values by difference $2k\pi$ so $ln(x+iy)=ln(x+i(y+2k\pi))$.

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