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Math Help - Principal branch

  1. #1
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    Principal branch

    We define the principal branch of the inverse tangent function by,

    Arctan(z) = \frac{i}{2}\log(\frac{i+z}{i-z})

    For which values of z is Arctan(z) defined?
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  2. #2
    MHF Contributor chisigma's Avatar
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    The function \tan^{-1} z = \frac{i}{2} \ln \frac{i+z}{i-z} is defined for all  z \in \mathbb{C} with the only exceptions of the points z=i and z=-i, that represent the 'singularities' of the function ...

    Kind regards

    \chi \sigma
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Richmond View Post
    We define the principal branch of the inverse tangent function by,

    Arctan(z) = \frac{i}{2}\log(\frac{i+z}{i-z})

    For which values of z is Arctan(z) defined?
    Its defined for any value that doesn't cause \log\left(\frac{i+z}{i-z}\right) to be undefined.

    Can you try to take it from here?

    EDIT: chisigma beat me...
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  4. #4
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    So as long as \frac{i}{2} \log \frac{i+z}{i-z} is defined,
    we can say that Arctan(z) is defined too which are all values except  z=i and z=-i
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  5. #5
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    Quote Originally Posted by Richmond View Post
    Thanks

    So as long as \frac{i}{2} \log \frac{i+z}{i-z} is defined,
    we can say that Arctan(z) is defined too which are all values except  z=i and z=-i
    No, for beeing log z well defined, and analytic, we need eliminate a branch
    (line beginning from 0). So, if nothing else is especified, logz is defined in
    C\[0,\infty). You need to solve the equatiom

    \arg\frac{i+z}{i-z}=\pi
    to obtain the correct domain of definition.
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