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$\newcommand{\Ln}{\operatorname{Ln}}$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 $5i\pi/2$ ?

majid azimi
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    Ah, you've discovered the wonderful world of complex multi-valued functions! https://en.wikipedia.org/wiki/Multivalued_function ; https://en.wikipedia.org/wiki/Riemann_surface ; https://en.wikipedia.org/wiki/Complex_logarithm –  Aug 18 '21 at 22:23
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    Although $5i\pi/2 \neq i\pi/2,~$ you have that $e^{5i\pi/2} = e^{i\pi/2}.$ You have the Real Analysis result that for $a,b,c \in \Bbb{R}, ~a > 0,~a^b = a^c \iff b = c.$ The Complex Analysis analog of the Real Analysis result is false. – user2661923 Aug 18 '21 at 22:27
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    That is because the complex logarithm is kinda complicated. If you interpret $i$ as a $90$ degree rotation, this can be explained by the jump discontinuity of the complex logarithm at the negative real axis, there it does a $2\pi i$-valued jump. – HyperPro Aug 18 '21 at 22:31
  • Wow! Very complex and very very Attractive. Thanks for your answers. Is there any more documents about this subject? – majid azimi Aug 18 '21 at 22:39
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    Simply put: the logarithm law does not hold for complex numbers. – TonyK Aug 18 '21 at 22:44
  • Why the logarithm law dosen't hold for complex numbers? is not this a big problem in mathematics? I can't understand this bug :( is this the bug? Or is this provable and it has a absolutely logical proof? – majid azimi Aug 18 '21 at 22:56
  • It's not a big problem, I'd argue that it lowers the "value" of the $\Ln$ function in mathematics, that's why you rarely see it being used (at least that's my experience). – kingW3 Aug 18 '21 at 23:07
  • "Why the logarithm law dosen't hold for complex numbers?" It seems to me that you have discovered the reason for this by yourself. It's exactly what your question is about, isn't it? – TonyK Aug 19 '21 at 16:51

2 Answers2

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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 $2\pi.$

Michael Hardy
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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.

Oddly Asymmetric
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