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Math Help - Inverse of a complex function

  1. #1
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    Inverse of a complex function

    hi, i have been set the following problem, where z is a complex number:


    What is the domain of the function f(z) = (3z+1)/(z+i) ?

    Prove that f, defined in the domain of f, has an inverse function (f-1),

    i.e. check all necessary properties for the existence of an inverse function.

    Determine f-1 and its domain.

    I have managed to find the domain and image of the function, and therefore the domain of f-1,

    but the problem im having is showing that the function is 1-1 (or injective), in order to show that it has an inverse.

    any help would be greatly appreciated!
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  2. #2
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    Re: Inverse of a complex function

    Straight algebra and the definition works:

    \text{If }z_1, z_2 \in \mathbb{C} \backslash \{-i\}, \text{ and } f(z_1) = f(z_2), \text{ then }

    \frac{3z_1 + 1}{z_1 + i} = \frac{3z_2 + 1}{z_2 + i}, \text{ so }

    (3z_1 + 1)(z_2 + i) = (3z_2 + 1)(z_1 + i), \text{ so }

    3z_1z_2 + z_2 +3iz_1 + i = 3z_1z_2 + z_1 +3iz_2 + i, \text{ so }

    z_2 +3iz_1 = z_1 +3iz_2, \text{ so }

    z_2-z_1 = 3i(z_2-z_1), \text{ so }

    0 = (-1+3i)(z_2-z_1), \text{ so }

    0 = z_2-z_1, \text{ so }

    z_1 = z_2. \text{ Thus f is injective on its domain.}
    Thanks from paulluap1991
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  3. #3
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    Re: Inverse of a complex function

    This message looks very similar to the coursework I set my Complex Analysis class. Paul please come see me in my office when I return. This is a clear violation of the academic code of conduct.
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