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Math Help - Simplifying Complex Numbers (i)

  1. #1
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    Simplifying Complex Numbers (i)

    Okay, I think I did this right, but I'm going to post it here just to make sure.

    Simplify the complex number i^31 as much as possible.

    Here's what I did:

    i^31
    =i^30+1
    =i^6*5+1
    =i^1
    =i


    So would my final answer just be i? (assuming I did that right)
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  2. #2
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    Hello,

    The method is correct, but not the result :

    i=i
    iČ=-1
    i^3 = -i
    i^4 = 1

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  3. #3
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    I don't understand how the result is incorrect.

    Shouldn't it be simply i?


    What isn't right?
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  4. #4
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    =i^6*5+1
    =i^1
    This step is strange.

    This means you suppose i^(6*5) = 1, which is false.

    You may write that i^{31}=i^{28}i^3 = (i^4)^7 i^2 i

    i^4 = 1 as i've shown you.

    So what's the result ?
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  5. #5
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    So would it be (1^7)i ?
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  6. #6
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    Learn these four numbers i,\,i^2  =  - 1,\,i^3  =  - i,\,i^4  = 1.
    Now divide the exponent by 4 and take the remainder.
    Thus i^{31}  = i^3  =  - i because 31 divided by 4 leaves a remainder of 3.
    Here why that works.
    i^{31}  = i^{28 + 3}  = \left( {i^4 } \right)^7 \left( {i^3 } \right) = \left( 1 \right)^7 \left( {i^3 } \right) =  - i
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  7. #7
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    Quote Originally Posted by eraser851 View Post
    So would it be (1^7)i ?
    The very first property of i is that iČ=-1
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  8. #8
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    eraser,

    Plato is right. When dealing with Imaginary numbers, learn those 4 powers. If the power you are to raise i to exceeds 4, you want to use i^4 and see what the remainder of powers is after you factor out a 4. Either 4 divides in evenly or it doesn't. If it does not, you are left with either i^1, i^2, or i^3.

    See here:

    Imaginary Numbers
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