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Thread: proving complex stuff

  1. #1
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    proving complex stuff

    if A+iB = (a.+ib.) (a,+ib,), prove (a.^2+b.^2) (a,^2 +b,^2 ) = (A^2 + B^2)
    a. & b. is suppose to be subcript 1 and a, & b, is suppose to be subscript 2.

    i dont see any short way of proving this, i have tried squaring A+iB and re-arranging to prove (A^2 + B^2) but hit a dead end.

    any suggestions on how to attemp?
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  2. #2
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    Quote Originally Posted by purebladeknight View Post
    if A+iB = (a.+ib.) (a,+ib,), prove (a.^2+b.^2) (a,^2 +b,^2 ) = (A^2 + B^2)
    a. & b. is suppose to be subcript 1 and a, & b, is suppose to be subscript 2.

    i dont see any short way of proving this, i have tried squaring A+iB and re-arranging to prove (A^2 + B^2) but hit a dead end.

    any suggestions on how to attemp?
    $\displaystyle A + iB = (a_1 + ib_1)(a_2 + ib_2)$

    $\displaystyle = a_1a_2 + ia_1b_2 + ia_2b_1 + i^2b_1b_2$

    $\displaystyle = a_1a_2 - b_1b_2 + i(a_1b_2 + a_2b_1)$


    Therefore $\displaystyle A = a_1a_2 - b_1b_2$ and $\displaystyle B = a_1b_2 + a_2b_1$.


    Now let's try to prove the statement that $\displaystyle (a_1^2 + b_1^2)(a_2^2 + b_2^2) = A^2 + B^2$.


    $\displaystyle LHS = (a_1^2 + b_1^2)(a_2^2 + b_2^2)$

    $\displaystyle = (a_1a_2)^2 + (a_1b_2)^2 + (a_2b_1)^2 + (b_1b_2)^2$.


    $\displaystyle RHS = A^2 + B^2$

    $\displaystyle = (a_1a_2 - b_1b_2)^2 + (a_1b_2 + a_2b_1)^2$

    $\displaystyle = (a_1a_2)^2 - 2a_1a_2b_1b_2 + (b_1b_2)^2 + (a_1b_2)^2 + 2a_1a_2b_1b_2 + (a_2b_1)^2$

    $\displaystyle = (a_1a_2)^2 + (a_1b_2)^2 + (a_2b_1)^2 + (b_1b_2)^2$

    $\displaystyle = LHS$.


    Q.E.D.
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