Question

How do you simplify i^40

Answer 1

$\sqrt{-1} \ \ \ | (...)^{2} \\ \\ i^{2} = -1 \\ \\ i^{3} = \sqrt{-1}^{3} = ( \sqrt{-1})^{2}* \sqrt{-1} =-i \\ \\ i^{4}=(i^{2})^{2}=(-1)^{2}=1 \\ \\ i^{40} = (i^{4})^{10} = 1^{10} = 1$

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Answer 2

Well
i=√-1 so
i^2=(√-1)^2=-1
remember that
x^(mn)=(x^n)^m
i^40=(i^2)^20
we know that i^2=-1 so
i^40=-1^20
-1^2=1
-1^3=-1
-1^4=1
-1^5=-1
we conclude that even powers of -1 yeild +1 so the answer is +1 (positive 1)