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Determine if $\operatorname{Log}(zw)=\operatorname{Log}(z) + \operatorname{Log}(w)$.

I know that it is not equal but what's wrong with this proof ? We know that $$\operatorname{Log}(zw) = \operatorname{Log}(r_1r_2e^{i(\alpha_1+\alpha_2)})$$ By definition of the main branch of $\operatorname{Log}(z)$, we get the main branch of $\operatorname{Arg}(z)$: $$\ln(r_1r_2) + i \operatorname{Arg}(e^{i(\alpha_1+\alpha_2)}) = \ln{r_1r_2} + i (\alpha_1+\alpha_2)$$ Now $$\operatorname{Log}(r_1e^{i\alpha_1}) + \operatorname{Log}(r_2e^{i\alpha_2}) = \ln{r_1}+i\operatorname{Arg}(e^{i\alpha_1})+\ln{r_2}+i\operatorname{Arg}(e^{i\alpha_2}) = \ln{r_1r_2}+i(\alpha_1+\alpha_2)$$

Gary
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amit
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    Notice Log is not a function on $\mathbb{C}\setminus{0}$ but a relation, that is for each $z\neq0$ there are are several (in fact countably many) $Log(z)$, they all differ by an integer multiple of $2\pi i$. Even when you set a brach of log, the sum $\alpha_1+\alpha_2$ may exceed $2\pi $. – Mittens Feb 05 '24 at 17:27
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    Cf. https://math.stackexchange.com/questions/642568/complex-logarithm-logzw-neq-logz-logw – Travis Willse Feb 05 '24 at 17:59
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    @Mittens In some conventions, $\log$ is the multi-valued relation and $\mathrm{Log}$ is function that takes the principal value of the argument. Your final sentence is the key to the issue though. – ConMan Feb 06 '24 at 00:47
  • @ConMan: I agreed with you. Still may argument at the end holds. – Mittens Feb 06 '24 at 00:48

1 Answers1

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If we are taking $\mathrm{Log}$ on the principal branch as you say, then we know that $\mathrm{Arg}(\mathrm{Log}(z)) \in (-\pi, \pi]$. So the reason that the proof fails is because you can have, for example, $\mathrm{Arg}(\mathrm{Log}(z)) = \mathrm{Arg}(\mathrm{Log}(w)) = \pi$, so you then get

$$\begin{eqnarray} \mathrm{Arg}(\mathrm{Log}(zw)) & = & 0 \\ & \neq & 2 \pi & = & \mathrm{Arg}(\mathrm{Log}(z)) + \mathrm{Arg}(\mathrm{Log}(w)) \end{eqnarray}$$

ConMan
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