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Logarithms of negative numbers must be complex.

But how do you find $\ln{(-2)}$ expressed in something like $x \cdot i$ where $x \in \mathbb{R}$?

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    Me too, thanks! You can put it as an answer.2011-12-20

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First note that allowing complex values, $\ln$ stops being unique (or, alternatively, it's not defined on all of $\mathbb{C}={0}$).

The key property of $\ln$ is that $e^{\ln z} = z$ for all complex numbers $z\neq 0$.

Write $\ln(z) = u(z) + iv(z)$ for 2 real valued functions $u(z)$ and $v(Z)$. Then the key property gives $e^{u(z)}[\cos v(z) + i\sin v(z)] = z$

In particular, we must have $|z| = e^{u(z)}$ so $u(z) = \ln(|z|)$ (this is the usual real valued $\ln$ defined on positive real numbers).

Likewise, we must have $v(z)= arg(z) + 2\pi k$ for some $k\in\mathbb{Z}$.

Putting this together gives $\ln(z) = \ln(|z|) + i(arg(z) + 2\pi k)$, valid for any nonzero complex number.

Setting $z = -2$, we have $|z| = 2$ and $arg(-2) = \pi$ so $\ln(-2) = \ln(2) + i(\pi + 2\pi k)$.

Just to verify this works, we get $e^{\ln(-2)} = e^{\ln(2) + i(\pi + 2\pi k)} = 2e^{i\pi} = -2.$

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    Or, alternatively again, the logarithm can be unique _and_ defined on all of $\mathbb C\setminus\{0\}$, only not continuous on all of $\mathbb C\setminus\{0\}$.2011-12-20