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Is This True For Complex Numbers?

  1. Nov 29, 2004 #1
    i^57 is simplified to i ?
  2. jcsd
  3. Nov 29, 2004 #2
    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...


    (Don't be mad if I am completely wrong though, its just what I remember)
  4. Nov 29, 2004 #3
    Who is correct lol? i or -i? which one?

    OH NO!! NOW im not sure well 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?
  5. Nov 29, 2004 #4
    i have to type some stuff to make my message longer
    answer is:

  6. Nov 29, 2004 #5


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    This is correct.
    [tex]i^{57} = i^{(56+1)} = i^{56}*i = (i^4)^{14}*i = 1^{14}*i = 1*i = i [/tex]
    Last edited: Nov 29, 2004
  7. Nov 29, 2004 #6
    Oh, so there has to be a remainder of 1, guess I forgot that.

  8. Nov 30, 2004 #7

    WOW GOKU ur answer is COMPLEX!! lol
    holy made me think! A simple question but a long way of simplifying it. Thanks soooo much yayay I got it right. Thanks everyone else for ur help! :tongue2:
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