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Thread: [SOLVED] Image of a complex function

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Image of a complex function

    Let $\displaystyle f(z)=z+ \frac{1}{z}$.
    What's the image of :
    1)A circle with radius $\displaystyle r$ centered at $\displaystyle z=0$?
    2)A circular semi-ring $\displaystyle \{ z=re^{i\theta} | 0 \leq \theta \leq \pi$ , $\displaystyle 1 \leq r \leq b \}$, $\displaystyle b \geq 1$?

    What's the preimage by $\displaystyle f$ of the semi-plane $\displaystyle \{ w : w \geq 0 \}$?

    -----------------------------------------------------------------
    For now I'm only looking for help for the 1).
    My attempt : I wrote $\displaystyle z$ as $\displaystyle re^{i \theta}$ to get $\displaystyle f(re^{i\theta})=r (\cos \theta + i \sin (\theta))+ r^{-1} (\cos \theta + i \sin \theta)^{-1}$. I don't recognize what it is. Especially the part $\displaystyle r^{-1} (\cos \theta + i \sin \theta)^{-1}$.
    Any idea about how I could rewrite $\displaystyle f(re^{i\theta})$?
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    Quote Originally Posted by arbolis View Post
    Let $\displaystyle f(z)=z+ \frac{1}{z}$.
    What's the image of :
    1)A circle with radius $\displaystyle r$ centered at $\displaystyle z=0$?
    2)A circular semi-ring $\displaystyle \{ z=re^{i\theta} | 0 \leq \theta \leq \pi$ , $\displaystyle 1 \leq r \leq b \}$, $\displaystyle b \geq 1$?
    Hint: This Mobius trasformation maps $\displaystyle |z|=r$ into another circle.

    What's the preimage by $\displaystyle f$ of the semi-plane $\displaystyle \{ w : w \geq 0 \}$?
    Let $\displaystyle V$ be this set.

    Find $\displaystyle f^{-1}$ (it exists), and note that $\displaystyle f^{-1}[V] = f^{-1}[V]$ where LHS is the inverse-image under $\displaystyle f$ and RHS is the image under $\displaystyle f^{-1}$.
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Hint: This Mobius trasformation maps $\displaystyle |z|=r$ into another circle.
    How strange. By the way I never learned about Mobius transformations yet but I know I will in less than a month I guess.


    Quote Originally Posted by TPH
    Let $\displaystyle V$ be this set.

    Find $\displaystyle f^{-1}$ (it exists), and note that $\displaystyle f^{-1}[V] = f^{-1}[V]$ where LHS is the inverse-image under $\displaystyle f$ and RHS is the image under $\displaystyle f^{-1}$.
    How would I find the inverse function? I'm thinking about finding a matrix and finding the inverse of it... but I'm unsure. I'd like to know how you would do it. (without posting the full solution )
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    Quote Originally Posted by arbolis View Post
    How strange. By the way I never learned about Mobius transformations yet but I know I will in less than a month I guess.



    How would I find the inverse function? I'm thinking about finding a matrix and finding the inverse of it... but I'm unsure. I'd like to know how you would do it. (without posting the full solution )
    Let $\displaystyle g(z)$ be the inverse function, it means, $\displaystyle f(g(z)) = z$, therefore, $\displaystyle g(z) + \tfrac{1}{g(z)} = z$.
    What does that make $\displaystyle g(z)=?$
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let $\displaystyle g(z)$ be the inverse function, it means, $\displaystyle f(g(z)) = z$, therefore, $\displaystyle g(z) + \tfrac{1}{g(z)} = z$.
    What does that make $\displaystyle g(z)=?$
    Note sure I didn't make any mistake : $\displaystyle g(z)=\frac{z \pm \sqrt{z^2-4}}{2}$.
    I should check it out.

    Edit : Indeed I made an error. I'll try it again.
    Edit: Ah no! It is right, at least $\displaystyle g(z)=\frac{z + \sqrt{z^2-4}}{2}$ works...
    Thanks TPH, I'll try what comes next.
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    I made a mistake, I wrote "this Mobius transformation", but it is not a Mobius transformation!
    So it is not necessary true that circles go to circles. Be careful in first part.
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  7. #7
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I made a mistake, I wrote "this Mobius transformation", but it is not a Mobius transformation!
    So it is not necessary true that circles go to circles. Be careful in first part.
    Ah ok! Thanks for the correction.
    It seemed strange to me because $\displaystyle f(z)=z$ would send circles to exactly the same circles, but if you add the $\displaystyle \frac{1}{z}$ term, it shouldn't be circles anymore.
    A circle is mapped to the same circle plus the inverse of this circle.
    I don't know what is this "inverse of this circle".
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  8. #8
    MHF Contributor arbolis's Avatar
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    Hey TPH if you mind, 1) is an ellipse.
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