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Math Help - cosine inverse

  1. #1
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    cosine inverse

    Solve the following equation:

    \cos z = 3 - 2i.

    My book says that

    \cos^{-1}z= -i \log[z+i(1-z^2)^{\frac{1}{2}}]. It also says that (1-z^2)^{\frac{1}{2}} is a double valued fuction. How do I use this formula or is there a better way to do this? I don't see how to get the double value it is talking about.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by xboxlive89128 View Post
    Solve the following equation:

    \cos z = 3 - 2i.

    My book says that

    \cos^{-1}z= -i \log[z+i(1-z^2)^{\frac{1}{2}}]. It also says that (1-z^2)^{\frac{1}{2}} is a double valued fuction. How do I use this formula or is there a better way to do this? I don't see how to get the double value it is talking about.

    let

    x = \cos ^{-1} z x here present cos inverse of z we want to find x value without using cos function , x is a complex number

    so

    \cos x = z

    \frac{e^{ix}+e^{-ix}}{2} = z

    e^{ix} + e^{-ix} =2z

    e^{2ix} - 2z e^{ix} +1 = 0

    let v=e^{ix}

    v^2 -2z v +1 =0

    v=\frac{-(-2z)\mp \sqrt{4z^2 - 4}}{2}

    e^{ix} = \frac{2z \mp 2\sqrt{z^2-1}}{2}

    e^{ix} = z\mp \sqrt{z^2-1}

    ix = \ln (z + \sqrt{z^2 -1 } )

    x = -i \ln (z+\sqrt{z^2-1})

    x = -i \ln (z+\sqrt{-1(1-z^2)}

    x = -i \ln (z +\sqrt{-1}(\sqrt{1-z^2}))

    x = -i \ln (z+i \sqrt{1-z^2})=\cos ^{-1} z End
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  3. #3
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    Quote Originally Posted by xboxlive89128 View Post
    I don't see how to get the double value it is talking about.
    Because the square root is multi-valued and defined as:

    \sqrt{z}=r^{1/2}e^{i/2(\theta+2k\pi)},\quad k=0,1

    You've seen that formula right? And also, the log function is infinitely valued:

    \log(z)=\ln r+i(\theta+2k\pi),\quad k=0,\pm 1, \pm 2,\cdots

    Therefore, cos^{-1}(z) is infinitely-valued.
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