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Algebra II

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Please help me with this problem.

Simplify.

√(-24 - 10i)

  • Algebra II -

    oh, and also, if it is not too much too ask, please include steps and explanations. ty

  • Algebra II -

    √(-24 - 10i)
    If you want to find a perfect square which equals -24-10i, it will have to include -5i, so

    (a-5i)^2 = -24-10i
    a^2 - 10ai - 25 = -24-10i
    Hmmm. Looks like a=1

    (1-5i)^2 = 1 - 10i - 25 = -24-10i
    So,
    √(-24 - 10i) = 1-5i or -1+5i

    Or, if you use polar form, you can see that
    r=26
    tanθ = 5/12
    the square root has
    r = √26
    tanθ = (1-(-12/13))/(-5/13) = -5
    so, we get the same result.

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