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**HallsofIvy** You cannot go from $\displaystyle e^{2ni\pi}= 1$ to $\displaystyle 2ni\pi= ln(1)= 0$ so quickly. In the complex numbers the logarithm is multi-valued. Specifically, if, in polar form, $\displaystyle x= re^{i\theta}$ then it is also true that $\displaystyle x= re^{i(\theta+ 2n\pi}$. $\displaystyle ln(x)= ln(r)+ 2n\pi$ where n can be any integer. In particular, since 1, in polar form, has r= 1, [itex]\theta= 0[/tex], ln(1) has many values, one of which is 0. The others are, just as you say, multiples of $\displaystyle 2\pi i$. It does not follow that, just because as a real function, ln(1)= 0 that $\displaystyle 2n\pi i= 0$.