Question

solve the iota power 68​

Answer 1

Answer:

$i^{68} = \sqrt{-1} ^{(68)}\\\sqrt{-1}^{2n} = -1\\68 = 2n , n = 34$

Step-by-step explanation:

iota to the power even number is always ( -1 )

iota to the power odd number is not ( -1 )

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