We know $\ln (-1)$ is $i\pi$ which can be found in the following way: $$ e^{i\pi} = \cos\pi \ + \ i\sin\pi = -1 \\ \implies \ln e^{i\pi} = \ln(-1)\implies \ln(-1) = i\pi $$ In the similar way, we can write: $$ e^{2i\pi} = \cos2\pi \ + \ i\sin2\pi = 1 \\ \implies \ln(e^{2i\pi}) = \ln(1) \implies 2i\pi = 0 $$ What is wrong here?
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7The complex $\log$ is a multi-valued function. – Evariste Jun 28 '21 at 10:49
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1When the input is a complex number, the function $f(z)=e^z$ is periodic with period $2\pi i$. So the complex logarithm is "multiple valued" i.e. can add $2 \pi i$ to anything and get the same logarithm. – coffeemath Jun 28 '21 at 10:50
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3For a discussion of the multi-valued-ness of the complex logarithm, see "Problems with inverting the complex exponential function" on Wikipedia. – Blue Jun 28 '21 at 10:53
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Unlike real numbers, non-zero complex numbers $z$ have infinitely many logarithms i.e. numbers $w$ such that $e^w=z$. The principal value generally accepted (denoted $\mathrm{Log}$) doesn't satisfy $\mathrm{Log}(e^a)=a$ for every $a\in\mathbb C$. Instead, for every $r\in\mathbb R_+^*$ and $\theta\in\mathbb R$, $$\mathrm{Log} (re^{i\theta})=\mathrm{ln}(r)+i(\theta \text{ mod } 2\pi)$$ with $(\theta \text{ mod } 2\pi)\in[-\pi,\pi)$.
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