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Math Help - imaginary numbers

  1. #1
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    imaginary numbers

    Hello! I don't know how to foil and add/subtract an imaginary numbers.

     x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i))

    thanks...
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  2. #2
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    Quote Originally Posted by Anemori View Post
    Hello! I don't know how to foil and add/subtract an imaginary numbers.

     x^2 (x+2i) (x-2i) (x - (3-i)) (x-(3+i))

    thanks...
    x^2(x+ 2i)(x-2i)[(x-3)+i][(x-3)-i]

    x^2(x^2 - 4i^2)[(x-3)^2 - i^2]

    note that i^2 = -1 ... finish.
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    Quote Originally Posted by skeeter View Post
    x^2(x+ 2i)(x-2i)[(x-3)+i][(x-3)-i]

    x^2(x^2 - 4i^2)[(x-3)^2 - i^2]

    note that i^2 = -1 ... finish.

    x^2(x^2 - 4i^2)[(x-3)^2 - i^2]

    x^2(x^2+4)(x^2-6x+10)

    x^2(x^4 - 6x^3+14x^2-24x+40)

    x^6 - 6x^5+14x^4-24x^3+40x^2

    This is what i get... i hope this is right...
    Last edited by Anemori; March 14th 2010 at 07:25 PM.
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    Quote Originally Posted by Anemori View Post
    x^2(x^2 - 4i^2)[(x-3)^2 - i^2]

    x^2(x^2+4)(x^2-6x+10)

    x^2(x^4 - 6x^3+14x^2-24x+40)

    x^6 - 6x^5+14x^4-24x^3+40x^2

    This is what i get... i hope this is right...
    I dont know if this is right...

    when i plug in back the factor from this polynomial, it doesn't go to zero.
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  5. #5
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    Quote Originally Posted by Anemori View Post
    I dont know if this is right...

    when i plug in back the factor from this polynomial, it doesn't go to zero.
    the zeros are x = \pm 2i , x = 3 \pm i , and x = 0

    ... are these the values what you "plugged" in for x ?
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    Quote Originally Posted by skeeter View Post
    the zeros are x = \pm 2i , x = 3 \pm i , and x = 0

    ... are these the values what you "plugged" in for x ?

    yes. but its not real zero. maybe im doing it wrong. can you show me one of them
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  7. #7
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    Quote Originally Posted by Anemori View Post
    yes. but its not real zero. maybe im doing it wrong. can you show me one of them
    f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2<br />

    f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2

    f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2

    f(-2i) = -64 + 192i + 224 - 192i - 160 =0
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  8. #8
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    Quote Originally Posted by skeeter View Post
    f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2<br />

    f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2

    f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2

    f(-2i) = -64 + 192i + 224 - 192i - 160 =0
    oh ok... maybe my calculator if messed up. im getting huge numbers heheheh. but it wouldnt make any difference if I use the exponents first before multiplying the outside paranthesis?
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    Quote Originally Posted by skeeter View Post
    f(x) = x^6 - 6x^5 + 14x^4 - 24x^3 + 40x^2<br />

    f(-2i) = (-2i)^6 - 6(-2i)^5 + 14(-2i)^4 - 24(-2i)^3 + 40(-2i)^2

    f(-2i) = 64i^6 + 192i^5 + 224i^4 + 192i^3 + 160i^2

    f(-2i) = -64 + 192i + 224 - 192i - 160 =0

    only -2i and 3-i will result of zero in the function. the rest are not zero.
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  10. #10
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    Quote Originally Posted by Anemori View Post
    only -2i and 3-i will result of zero in the function. the rest are not zero.
    is that so?

    f(2i) = (2i)^6 - 6(2i)^5 + 14(2i)^4 - 24(2i)^3 + 40(2i)^2<br />

    f(2i) = 64i^6 - 192i^5 + 224i^4 - 192i^3 + 160i^2

    f(2i) = -64 - 192i + 224 + 192i - 160 = 0


    ... and I guarantee you that f(3+i) = 0 ... complex zeros always come in conjugate pairs.
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  11. #11
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    There is enough information for the user to solve the problem. There is no need to carry on this discussion anymore.

    Thread closed.
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