Is it true that $\log(i^2)\neq2\log(i)$?

Question

I think that $\log(i^2)\neq2\log(i)$ because $\log(z)=\ln|z|+\arg(z)$. The LHS will give values $2\pi$ part while the RHS gives values $4\pi$ apart. But I saw this rule used in some textbooks. Did I make any mistakes?

Answer 1

The best approach is probably to resolve not to rely on equalities between such multi-valued expressions. The work that needs to go into determining exactly to which extent they are valid is generally not commensurate with the insights you can get out of using them.

But if you must: What holds is that $\log(zw)=\log z + \log w$, and therefore $\log(i^2) = \log i + \log i$. But the right-hand side of this is not the same as $2\log i$ because in $\log i+\log i$ you're supposed to add every value of the left $\log i$ to every value of the right $\log i$, and not all possible outcomes of that can arise as a single value of $\log i$ added to itself.