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Math Help - complex simplifying

  1. #1
    Senior Member DivideBy0's Avatar
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    complex simplifying

    Simplify \sqrt{15+8i}.

    I got 4+i or -4 - i, but only one of these solutions is correct.

    Obviously \sqrt{15+8i}>0, but with my answers, wouldn't the pesky i term disable you from checking whether the expression is greater or less than 0? After all, i is not on the number line... I dunno, it just feels risky dealing with that i term.

    thanks
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  2. #2
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    Quote Originally Posted by DivideBy0 View Post
    Simplify \sqrt{15+8i}.

    I got 4+i or -4 - i, but only one of these solutions is correct.

    Obviously \sqrt{15+8i}>0, but with my answers, wouldn't the pesky i term disable you from checking whether the expression is greater or less than 0? After all, i is not on the number line... I dunno, it just feels risky dealing with that i term.

    thanks
    Given any (non-zero) number z its square root \sqrt{z} is \sqrt{|z|} \cdot e^{\theta i/2} where \theta is the angle it makes with the x-axis measure (-\pi,\pi].*

    So if z=15+8i then |z| = \sqrt{15^2+8^2} = \sqrt{298}.

    The angle that 15+8i makes with the x-axis is \theta = \tan^{-1} (15/8).

    So,
    e^{i\cdot \left( \frac{\tan^{-1}(15/8)}{2} \right) } = \cos \left( \frac{\tan^{-1}(15/8)}{2} \right) + i \sin \left( \frac{\tan^{-1} (15/8)}{2} \right)

    Use half-angle formulas,
    \frac{1}{\sqrt{2}} \sqrt{ 1 + \cos \tan^{-1}(15/8) } + i\frac{1}{\sqrt{2}} \sqrt{ 1 - \cos \tan^{-1}(15/8)}

    Can you simplify that?

    *)And this is computed with Euler formula: e^{xi} = \cos x + i\sin x.
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