- #1

- 479

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##[cosθ + isinθ]^n = [cos(θ + 2kπ) + isin(θ + 2kπ)]^n##

##cos(nθ) + isin(nθ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)##

##cos(nθ + 2mπ) + isin(nθ + 2mπ) = cos(nθ + 2nkπ) + isin(nθ + 2nkπ)## for ##m ∈ ℤ##

Now consider the special case ##n = 1/p## for ##p ∈ ℤ##

##cos(\frac{θ}{p} + 2mπ) + isin(\frac{θ}{p} + 2mπ) = cos(\frac{θ}{p} + \frac{2kπ}{p}) + isin(\frac{θ}{p} + \frac{2kπ}{p})##

Is this a contradiction?