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Comment by twiss

That's not his argument; he says defining Θ = e^2πi is not useful because e^2πi = 1, so Θ and Θ^x would also be 1. That's why he defined Θ = e^2π instead, so that Θ^x (or possibly Θ^ix) is a useful operation.
parent twiss 3 years ago

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