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This is a proof I made a year ago and at that time, I didn't see any problems with it. Could anyone point out what is wrong here?

Consider the following expresion: $(-1)^{(4n+3)/2}$, where $n \in \Bbb Z_+$. We have that: \begin{align} (-1)^{(4n+3)/2} &= [(-1)^{4n+3}]^{1/2} = (-1)^{1/2} = i \\ (-1)^{(4n+3)/2} &= [(-1)^{1/2}]^{4n+3} = i^{4n+3} = -i \end{align} That means $i=-i$ $\implies$ $i=0$.

Lê Thành Đạt
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1 Answers1

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A 'square root' is not a function in complex numbers - there are two distinct values of $\sqrt{-1},$ and if you take the other one, which is $-i$, the result would be $-i=-i.$

CiaPan
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