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Math Help - Inverse Functions & Complex Numbers

  1. #1
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    Inverse Functions & Complex Numbers

    1.
    (Sorry if this is a little big.)

    I don't really know where to start. I substituted z = a + bi into the equation, but that didn't seem to help me.

    2. Let g:[b, 2] --> R, where g(x) = 1 - x2, if b is the smallest real value such that g has an inverse function, find b and g-1(x).

    I know that b = 0, but why? I can find the inverse function, but I can't quite grasp why 0 is the smallest value which allows g to have an inverse function.
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  2. #2
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    Re: Inverse Functions & Complex Numbers

    Quote Originally Posted by Fratricide View Post
    1.
    (Sorry if this is a little big.)

    I don't really know where to start. I substituted z = a + bi into the equation, but that didn't seem to help me.
    Try it again. That is the correct method.


    Quote Originally Posted by Fratricide View Post
    2. Let g:[b, 2] --> R, where g(x) = 1 - x2, if b is the smallest real value such that g has an inverse function, find b and g-1(x).

    I know that b = 0, but why? I can find the inverse function, but I can't quite grasp why 0 is the smallest value which allows g to have an inverse function.
    Because if b<0, then g(x) is no longer one-to-one. g(-x) = 1-(-x)^2 = 1-x^2 = g(x).
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  3. #3
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    Re: Inverse Functions & Complex Numbers

    hi Fratricide
    I will offer half an appology to start with because you probably know a lot more than me
    but
    with the first one couldn't you just substitute in the value of z and then multiply top and bottom by the conjugate of the denominator
    then put the real part=0 and the coefficient of i = 1 ?
    I did it and I think it worked but I might have made a mistake.
    Melody.
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  4. #4
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    Re: Inverse Functions & Complex Numbers

    Quote Originally Posted by Melody2 View Post
    hi Fratricide
    I will offer half an appology to start with because you probably know a lot more than me
    but
    with the first one couldn't you just substitute in the value of z and then multiply top and bottom by the conjugate of the denominator
    then put the real part=0 and the coefficient of i = 1 ?
    I did it and I think it worked but I might have made a mistake.
    Melody.
    Another method: \dfrac{z+i}{z+2} = i implies z+i = (z+2)i. Plug in z = a+bi, multiply out, and equate coefficients.
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  5. #5
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    Re: Inverse Functions & Complex Numbers

    Quote Originally Posted by Fratricide View Post
    1.
    (Sorry if this is a little big.)

    I don't really know where to start. I substituted z = a + bi into the equation, but that didn't seem to help me.

    2. Let g:[b, 2] --> R, where g(x) = 1 - x2, if b is the smallest real value such that g has an inverse function, find b and g-1(x).

    I know that b = 0, but why? I can find the inverse function, but I can't quite grasp why 0 is the smallest value which allows g to have an inverse function.
    There's no need to plug in \displaystyle \begin{align*} z = a + i\,b \end{align*} at all...

    \displaystyle \begin{align*} \frac{z + i}{z + 2} &= i \\ z + 1 &= i\left( z + 2 \right) \\ z + 1 &= i\,z + 2i \\ z - i\,z &= -1 + 2i \\ z \left( 1 - i \right) &= -1 + 2i \\ z &= \frac{-1 + 2i}{1 - i} \\ z &= \frac{ \left( -1 + 2i \right) \left( 1 + i \right) }{ \left( 1 - i \right) \left( 1 + i \right) } \\ z &= \frac{ -1 - i + 2i - 2}{1^2 + 1^2} \\ z &= -\frac{3}{2} + \frac{1}{2}i  \end{align*}
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  6. #6
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    Re: Inverse Functions & Complex Numbers

    Hi SlipEternal
    Thanks, that makes sense to me. Only trouble is I get 2 different answers. I'll take a break and then try to see where my stupid mistake lies.
    Melody.
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    Re: Inverse Functions & Complex Numbers

    Thanks for the help.
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