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Posted by on Saturday, October 27, 2012 at 8:31pm.


  • College math - , Sunday, October 28, 2012 at 12:23am

    All these complex value problems can be treated just like polynomials, where the reals and imaginaries have to be treated separately

    -8i(2i) - 8i(-7)
    -16i^2 + 56i
    16 + 56i

    Don't forget that i^2 = -1
    also key: (a+bi)(a-bi) = a^2 - b^2i^2 = a^2+b^2

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