$i^5=i$ so $\operatorname{Ln}(i^5)$ must be equal to $\operatorname{Ln}i$ but why ... $\newcommand{\Ln}{\operatorname{Ln}}$We know that $i^5=i$
So we must be have $\Ln(i^5)=\Ln i$
Now, according to the "$\Ln$ Law" we must be have: $\Ln(i^5)=5\Ln i$
We will also know that $\Ln i = i\pi/2$
and
$5\Ln i=5i\pi/2$
So, why $i\pi/2$ is not equal to $5i\pi/2$ ?
 A: Some people suffer horrible pain when one uses the term "multiple-valued function", but I'm not sure they're right. Certainly they are right if one construes "function" in a certain way that has become standard and conventional. So I have doubts about some standards and conventions.
Consider $\text{“}\pm\sqrt{2}\,\text{.”}$ There you have the double-valued square-root function. There's also quadruple-valued fourth-root function. The numbers $\pm1,\pm i$ are fourth roots of $1.$
Likewise $\operatorname{Ln}(e^r(\cos\theta+i\sin\theta)) = r+i\theta$ is multiple-valued since $\theta$ can be any of infinitely many numbers differing from each other by integer multiples of $2\pi.$
A: It is simply a matter that log is not smoothly varying at 0 (because there's a big singularity there.) Think about its taylor expansion, it's centred around $x=1$ instead of $x=0$. That's because $\log$ is well behaved in a disc around 1 in the complex plane. Because this singular behaviour, you have to cut along the complex plane, as though you're cutting into the paper, tearing a bit of it apart. You cut it along the negative imaginary axis.
Now when you have $z^a$, the complex number $z$ is rotating around the origin some number of times (and scaled some amount). It passes through this cut a few times, so takes on a different log value as a result. Now when you apply $\log(z^a)=a\log(z)$,you're essentially ignoring the rotation, and just looking at the scaling. You say that the scaling from exponentiation is the same as scaling by multiplying by a on the right. But you're completely ignoring the rotation that happens to $z^a$.
For that reason $\log(z^a)=a\log(z)$ works for the positive real numbers, because there is no rotation when you exponentiate a real number to another real number, but fails to be true for other complex values- since they have non-zero rotation.
