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Math Help - Help complex numbers

  1. #1
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    Help 1 ???

    Perform the indicated operations: (2+5i)(1-3i) all over 5-2i
    Last edited by Brooke; July 6th 2006 at 03:42 PM. Reason: not shown correctly
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  2. #2
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    Hello, Brooke!

    I assume you can do basic Algebra and that you know that i^2 = -1


    Perform the indicated operations: \frac{(2+5i)(1-3i)}{5-2i}

    Simplify the numerator: . (2 + 5i)(1 - 3i) \:=\:2 - 6i + 5i - 15i^2

    . . and we have: . 2 - i -15(-1)\:=\:2 - i + 15 \:= \:17 - i

    The problem becomes: . \frac{17 - i}{5 - 2i}


    Now we are expected to rationalize the denominator.

    Multiply top and bottom by the conjugate of the denominator:

    . . \frac{17 - i}{5 - 2i}\cdot\frac{5 + 2i}{5 + 2i} \;= \;\frac{85 + 34i - 5i - 2i^2}{25 + 10i - 10i - 4u^2} \;= \frac{85 + 29i - 2(-1)}{25 - 4(-1)}

    . . = \;\frac{85 + 29i + 2}{25 + 4}\;=\;\frac{87+29i}{29} \;= \frac{\not{29}(3 + i)}{\not{29}} \;= \;3 + i

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