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Does 2ιπ=0?
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If, using Euler's equation, eιθ= cos(θ) ιsin(θ),
If you plug in 2π for theta, then you get e2ιπ=1
If you ln both sides, you get 2ιπ=0
How can this be?
If you then divide both sides by 2ιπ, you get 1=0, so this can't be true, but then how does the equation come out true?
I have proven Euler's equation by Laplace transforms, so I don't need a proof.
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- 11 years ago
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