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$${e}^{iz} = \cos(z)+i\sin(z)$$ and $$e^{i\pi}=-1$$ But then $$\ln(-1)$$ can be infinite many numbers (positive and negative), as $z$ is the natural logarithm of that number and the solution to the initial equation. My doubt: Is $$\ln(-1) = \pm (2k-1)i\pi$$ where $k$ is an integer, true? Is correct saying $\ln(-1)$ is, for example, equal to $-3i\pi$?

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up vote 2 down vote accepted

The logarithm can just be chosen branchwise. That means that you have to fix one $k$. Of course it will not extend to a continuous function on $\mathbb{C} -0$ exactly because running around $0$ one time adds an imaginary angle.

In particular the complex $e^{z}$ has no inverse, for all $z \in \mathbb{C}-0$.

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The logarithm is what is known as a "multivalued function" in complex analysis. That is, $\log(z)$ is not just one complex number, it is the set of all $w$ such that $e^w = z$. If $w_0$ is one of these, then these are $w_0 + 2 \pi i n$ for all integers $n$. For many purposes we would like to single out one particular choice: a "branch" of the logarithm function. A popular choice (the "principal branch", often denoted as $\text{Log} (z)$), is given by $\text{Log}(r e^{i\theta}) = \ln(r) + i \theta$ for $r > 0$ and $-\pi < \theta \le \pi$. So $\text{Log}(-1) = i \pi$, but all other odd integer multiples of $i \pi$ are also values of $\log(-1)$.

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