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Google returns that response, not sure why is a complex number.

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    $\begingroup$ As you know, it cannot be a real number. $\endgroup$
    – user228113
    Jun 20, 2017 at 7:33

2 Answers 2

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If you remember the famous 'Euler' formula:

$e^{\pi i}+1=0$

And assuming $\ln$ verify $\ln(\exp(z)) = z$, you get the result $\ln(-1)=\pi i$.

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    $\begingroup$ + period $2\pi i$ $\endgroup$
    – TStancek
    Jun 20, 2017 at 7:43
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Using the principal value of the complex logarithm:

$$\ln\left(\text{z}\right)=\ln\left|\text{z}\right|+\arg\left(\text{z}\right)i\space\space\space\to\space\space\space\ln\left(-1\right)=\ln\left|-1\right|+\arg\left(-1\right)i=0+\pi i=\pi i\tag1$$

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