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Math Help - [SOLVED] Multiplication with complex numbers

  1. #1
    Nrt
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    [SOLVED] Multiplication with complex numbers

    I need to derive this equation and get the below result.

    ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6

    ===> (a^2-iab-b^2)(a+ib)=a^3-ib^3

    Also a hint teacher gave us:

    replace

    a by a^2
    b by -ib^2

    I tried few things (like -b^2=-1.b^2=i^2b^2) to involve complex numbers in the first equation but im confused and have no clue what to do.
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    Quote Originally Posted by Nrt View Post
    I need to derive this equation and get the below result.

    ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6

    ===> (a^2-iab-b^2)(a+ib)=a^3-ib^3

    Also a hint teacher gave us:

    replace

    a by a^2
    b by -ib^2

    I tried few things (like -b^2=-1.b^2=i^2b^2) to involve complex numbers in the first equation but im confused and have no clue what to do.
    Hi Nrt,

    The first equation
    ((a^2-b^2)^2+a^2b^2)(a^2+b^2)=a^6+b^6
    is an identity-- it is true for all values of a and b.

    Now
    (a^2-b^2)^2+a^2b^2 = a^4 - a^2b^2 +b^4
    so
    (a^4 - a^2b^2 +b^4)(a^2+b^2)=a^6+b^6.

    Now let x = a^2 and y = b^2 / i, so b^2 = iy.

    Then the identity becomes
    (x^2 - x(iy) + (iy)^2)(x + iy) = x^3 + (iy)^3
    so
    (x^2 -i x y -y^2)(x+iy) = x^3 - i y^3.

    This is the result you were asked to show, except that I used x and y instead of a and b. It seems less confusing to me that way. (I get confused very easily). But you can go back and replace x and y with a and b if you prefer it that way.
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  3. #3
    Nrt
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    Much appreciated, thank you.
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