SOLUTION: How do I simplify..imaginary numbers i to the fifteenth. i to the thirtyfifth. i to the fivehundredth and twelve. thank you!

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Question 530241: How do I simplify..imaginary numbers
i to the fifteenth.
i to the thirtyfifth.
i to the fivehundredth and twelve.
thank you!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
All you need to know is:
i%5E2=-1
i%5E4=i%5E%282%2B2%29=i%5E2%2Ai%5E2=%28-1%29%28-1%29=1 and
i%5E3=i%5E%282%2B1%29=i%5E2%2Ai=%28-1%29%2Ai=-i
Beyond that, you need to know how to divide by 4 and find the remainder.
i%5E15=i%5E%284%2A3%2B3%29=%28i%5E4%29%5E3%2Ai%5E3=1%5E3%2Ai%5E3=1%2A%28-i%29=-i
(When you divide 15 by 4 the remainder is 3).
i%5E35=i%5E%284%2A8%2B3%29=%28i%5E4%29%5E8%2Ai%5E3=1%5E8%2Ai%5E3=1%2A%28-i%29=-i
(When you divide 35 by 4 the remainder is 3 too).
On the other hand
i%5E21=i%5E%284%2A5%2B1%29=%28i%5E4%29%5E5%2Ai=1%2Ai=i and
i%5E14=i%5E%284%2A3%2B2%29=%28i%5E4%29%5E3%2Ai%5E2=1%5E3%2A%28-1%29=-1
Do you get the idea?
By the way,
512=2%5E9=2%5E2%2A2%5E7=4%2A2%5E7=4%2A128
It's a multiple of 4. No remainder.