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We can't just say that on the proof, right? we need an algebraic proof for this, right?

https://en.m.wikipedia.org/wiki/Complex_logarithm#Branches_of_the_complex_logarithm
Wikipedia!

Thanks for the link @thomas5267

I think the problem is that you have to choose a branch of \(arg(z)\) or else it won't work.

We cannot choose a case "it won't work" to prove the whole thing "won't work". I think.

So, we can prove the principal branch doesn't work on S, right?

like what oldrin says, let z =-1, then Log z = log|z|+iarg z= i(pi + 2kpi)