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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    How would you go about proving that if a product of complex numbers is equal to zero then at least one of the numbers is zero? Seems simple, but I don't know where to start.

    Greatly appreciate any help. Thanks in advance.
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  2. #2
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    Hello, Joe!

    I devised a proof . . . but there must be a shorter way.


    Prove: if a product of complex numbers is equal to zero,
    . . . . . then at least one of the numbers is zero.

    We have two complex numbers, a + bi and c + di, and their product is zero.

    Then: . . . . (a + bi)(c + di) .= .0

    And: .(ac - bd) + (ad + bc)i .= .0


    Hence: .ac - bd .= .0 . [1]

    and: . . ad + bc .= .0 . [2]


    Square [1]: . ac - 2abcd + bd .= .0

    Square [2]: . ad + 2abcd + bc .= .0


    Add: . .ac + ad + bc + bd .= .0

    Factor: . a(c + d) + b(c + d) .= .0

    Factor: . . . . . . (a + b)(c + d) .= .0


    We have the product of two real numbers equal to zero.
    . . Hence, one of the factors must equal zero.

    Since a + b is nonnegative, a + b = 0 only if a = 0 and b = 0.

    Similarly, if c + d = 0, then c = 0 and d = 0.


    Therefore, one of the complex numbers, a + bi or c + di, must be zero.
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