this post was submitted on 24 Jul 2024
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This is true but does not follow from the preceding steps, specifically finding it to be equal to -1. You can obviously find it from i²=-1 but they didn't show that. I think they tried to equivocate this expression with the answer for e^iπ^ which you can't do, it doesn't follow because e^iπ^ and i¹⁰ = ln(-1)¹⁰/pi¹⁰ are different expressions and without external proof, could have different values.
If we know the values of ln(-1)¹⁰ and pi¹⁰ we hypothetically could calculate their divided result as -1 instead of using strict logic, but it is missing a few steps. Moreover logs of negative numbers just end up with an imaginary component anyway so there isn't really any progress to be made on that front. Typing ln(-1)¹⁰ into my scientific calculator just yields i¹⁰pi¹⁰, (I'm guessing stored rather than calculated? Maybe calculated with built in Euler) so the result of division is just i¹⁰ anyway and we're back where we started.
You can find the value of ln(-1)¹⁰ by examining the definition of ln(x): the result z satisfies eᶻ=x. For x=-1, that means the z that satisfies eᶻ=-1. Then we know z from euler's identity. Raise to the 10, and there's our answer. And like you pointed out, it's not a particularly helpful answer.