Question 233025
What is the solution to i^16+i^20+i^22? and how do you do it? Thank You
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i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1
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Note: Every 4th power of i is 1.
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Your problem:
i^16 = 1
i^20 = 1
i^22 = i^20*i^2 = 1*-1 = -1
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Adding you get 1+1-1 = 1
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Cheers,
Stan H.