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Thread: Stereographic projection

  1. #1
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    Stereographic projection

    Hl! I have problems with this exercises

    Let $\displaystyle \varepsilon$ the Stereographic projection of $\displaystyle \mathbb{S}^1- \{i \}$, that is


    $\displaystyle \varepsilon : \mathbb{S}^1 - \{i \} \longrightarrow{\mathbb{R}}$


    where


    $\displaystyle \varepsilon(z)=\displaystyle\frac{RE(z)}{1-Im(z)}$


    Find a formula for $\displaystyle \varepsilon^{-1} :\mathbb{R}\rightarrow{\mathbb{S}^1}- \{i \} $

    I tried to take the line that passes through i and a point on the real axis to see where the circle intersects but I can not find the formula


    Thanks
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  2. #2
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    Re: Stereographic projection

    The stereographic projection maps

    $\displaystyle i \to \infty $

    $\displaystyle -i\to 0$

    $\displaystyle 1\to 1$

    Therefore its inverse sends

    $\displaystyle \infty \to i$

    $\displaystyle 0 \to -i$

    $\displaystyle 1 \to 1$

    it's easy to write down the corresponding Moebius transformation
    Thanks from cristianoceli
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  3. #3
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    Re: Stereographic projection

    Quote Originally Posted by Idea View Post
    The stereographic projection maps

    $\displaystyle i \to \infty $

    $\displaystyle -i\to 0$

    $\displaystyle 1\to 1$

    Therefore its inverse sends

    $\displaystyle \infty \to i$

    $\displaystyle 0 \to -i$

    $\displaystyle 1 \to 1$

    it's easy to write down the corresponding Moebius transformation
    You can make the first correspondence , please
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  4. #4
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    Re: Stereographic projection

    $\displaystyle \varepsilon ^{-1}(u)=\frac{i u+1}{u+i}$

    $\displaystyle u\in \mathbb{R}$
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  5. #5
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    Re: Stereographic projection

    Quote Originally Posted by Idea View Post
    $\displaystyle \varepsilon ^{-1}(u)=\frac{i u+1}{u+i}$

    $\displaystyle u\in \mathbb{R}$
    Thanks

    but, why the domain is $\displaystyle \mathbb{R} $and the codomain is$\displaystyle \mathbb{S}^1- \{i \}$ ?
    Last edited by cristianoceli; Aug 18th 2018 at 07:46 AM.
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