aisha
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i^57 is simplified to i ?
Nonok said:i^2 = -1
i^3 = -i
i^4 = 1
i^5 = i
57 is divisible by 3. So, if I remember my calc class then it would be...
-i
(Don't be mad if I am completely wrong though, its just what I remember)
aisha said:I divided the exponent by 4 and got a remainder of 1 which made me think that the answer is simply i
hmmm can someone tell us who is right?
Gokul43201 said:This is correct.
i^{57} = i^{(56+1)} = i^{56}*i = (i^4)^{14}*i = 1^{14}*i = 1*i = i