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Thread: complex numbers

  1. #1
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    Smile complex numbers

    If z = 3x + 4i and w = 1 - 2i, calculate:

    (z - w)^2
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle z-w=3x-1+6i$

    $\displaystyle (a+bi)^2=a^2-b^2+2abi$

    Replace a with 3x-1 and b with 6.
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  3. #3
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    Quote Originally Posted by Joker37 View Post
    If z = 3x + 4i and w = 1 - 2i, calculate:

    (z - w)^2
    $\displaystyle z-w = (3x+4i) - (1 - 2i) = (3x-1) + 6i$

    $\displaystyle [(3x-1) + 6i]^2$

    expand and simplify
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  4. #4
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    Quote Originally Posted by skeeter View Post
    $\displaystyle z-w = (3x+4i) - (1 - 2i) = (3x-1) + 6i$

    $\displaystyle [(3x-1) + 6i]^2$

    expand and simplify
    (3x - 1 - 6i)(3x - 1 + 6i)?
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  5. #5
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    Quote Originally Posted by Joker37 View Post
    (3x - 1 - 6i)(3x - 1 + 6i)?
    No. (A + B)^2 = (A + B)(A + B). NOT (A + B)(A - B).

    So (3x - 1 - 6i)^2 = (3x - 1 - 6i)(3x - 1 - 6i).

    Virtually the same idea as here: http://www.mathhelpforum.com/math-he...x-numbers.html
    Last edited by mr fantastic; Mar 3rd 2009 at 01:56 AM.
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