Lösung.

$ \mbox{${\operatorname{Log}}(-2\mathrm{i}) = {\operatorname{Log}}(2) - \mathrm{i}\frac{\pi}{2}$}$.
$ \mbox{${\operatorname{Log}}((-2\mathrm{i})^3) = {\operatorname{Log}}(8\mathrm{i}) = 3\,\log 2 + \mathrm{i}\frac{\pi}{2}$}$.
Es fällt auf, daß hier $ \mbox{${\operatorname{Log}}(z^k) \neq k{\operatorname{Log}}(z)$}$ gilt.