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Math Help - complex inverse trig

  1. #1
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    complex inverse trig

    let \sin(z) = 3 what is z equally to?

    \arcsin(z) = -i\log[i z+\sqrt{1-z^2}]

    substituting z for 3 I get

    \arcsin(3) = -i\log[i 3+\sqrt{-8}]

    I'm trying to get it into the form \ln|x| +(x+2n)\pi i

    but what I'm confused about is the \sqrt{-8} which gives \sqrt{8} + \frac{1}{2}(\pi+2\pi n)i= \sqrt{8} + \left(\frac{\pi}{2} +\pi n \right)i where n= 0,1

    now I'm not sure if I should take the above value and substitute it so that I would get:

    \arcsin(3) = -i\log\left[i 3+\sqrt{8} + \left(\frac{\pi}{2}\right)i \right] and

    \arcsin(3) = -i\log\left[i 3+\sqrt{8} + \left(\frac{3\pi}{2}\right)i \right]

    for some reason this equation don't look right even, considering the fact that I have 2 possible values.
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  2. #2
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    Quote Originally Posted by lllll View Post
    let \sin(z) = 3 what is z equally to?

    \arcsin(z) = -i\log[i z+\sqrt{1-z^2}]

    substituting z for 3 I get

    \arcsin(3) = -i\log[i 3+\sqrt{-8}]

    I'm trying to get it into the form \ln|x| +(x+2n)\pi i

    but what I'm confused about is the \sqrt{-8} which gives \sqrt{8} + \frac{1}{2}(\pi+2\pi n)i= \sqrt{8} + \left(\frac{\pi}{2} +\pi n \right)i where n= 0,1

    now I'm not sure if I should take the above value and substitute it so that I would get:

    \arcsin(3) = -i\log\left[i 3+\sqrt{8} + \left(\frac{\pi}{2}\right)i \right] and

    \arcsin(3) = -i\log\left[i 3+\sqrt{8} + \left(\frac{3\pi}{2}\right)i \right]

    for some reason this equation don't look right even, considering the fact that I have 2 possible values.
    \ln (3 i + \sqrt{-8}) = \ln(3 i + i 2 \sqrt{2}) = \ln(i[3 + 2 \sqrt{2}]) = \ln (3 + 2 \sqrt{2}) + \frac{\pi}{2} i + 2 n \pi i where n is an integer.
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