Can anyone help me out with explaining why $\log(zw)\neq\log(z)+\log(w)$?
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The reason why $\log(ab) \ne \log(a)+\log(b)$ is because log is a multi-valued function. Just like $\sqrt x = \pm x^{1/2}$, ${\rm Log}(x) = log(x) +2\pi k i$ for any integer k. I.e. ${\rm Log}(x)$ is not one distinct value. To show why this gives us problems, consider one example; $$0=\log(1)=\log(-1*-1)\ne \log(-1)+\log(-1)=\pi i +\pi i = 2\pi i$$ $$\log(-1*-1) \ne \log(-1)+\log(-1)$$
Christoph
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Elie Bergman
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Put $z=-1$ and $w=i{{{{{}}}}}.$
Git Gud
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I'm not sure this answer addresses the OP... – DonAntonio Jan 18 '14 at 12:53
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@DonAntonio I am an anti-psychologist. To me this is an explanation. – Git Gud Jan 18 '14 at 12:56
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Hmmm...and I'm a skeptic...:) How does this even hint towards the "why" in the OP? – DonAntonio Jan 18 '14 at 12:57
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@DonAntonio ^_^ To an anti-psychologist the 'why' you refer to is merely psychological state of mind. A proof, no matter how little explanatory it might look, is as a good as an explanatory one, and I mean this not in mathematical acceptance, but in epistemological acceptance. – Git Gud Jan 18 '14 at 13:00
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@DonAntonio: IMO, this very clearly hints towards the "why" the OP asked -- any dissatisfaction is because a person didn't want an answer to the "why" the OP asked, but an answer to a different, unasked question. – Jan 18 '14 at 13:12
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I can't see how it is towards the why, @Hurkyl. About the dissatisfaction: I can't understand why you meant ("a person"?) – DonAntonio Jan 18 '14 at 13:15
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@DonAntonio I think Hurkyl means that upon computing both sides of the equality, one realises that they differ by $2\pi i$ and that hints towards something, though that wasn't my intent at all. – Git Gud Jan 18 '14 at 13:18
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@Git Gud: Then what was your intent? – The_Sympathizer Mar 22 '14 at 01:55
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@mike4ty4 My intent was that a counter example explains the 'why'. See my first comment in this answer. – Git Gud Mar 22 '14 at 09:53
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They will always be equal modulo $2\pi i$. The problem is that $\rm exp$ is $2\pi i$ periodic so you have to pick which logarithm you take, and you can't make a continuous choice on all of $\mathbb C^*$.
Christoph
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