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Math Help - [SOLVED] Image of a complex function

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

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

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

    -----------------------------------------------------------------
    For now I'm only looking for help for the 1).
    My attempt : I wrote z as re^{i \theta} to get 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 r^{-1} (\cos \theta + i \sin \theta)^{-1}.
    Any idea about how I could rewrite f(re^{i\theta})?
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    Quote Originally Posted by arbolis View Post
    Let f(z)=z+ \frac{1}{z}.
    What's the image of :
    1)A circle with radius r centered at z=0?
    2)A circular semi-ring \{ z=re^{i\theta} | 0 \leq \theta \leq \pi , 1 \leq r \leq b  \}, b \geq 1?
    Hint: This Mobius trasformation maps |z|=r into another circle.

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

    Find f^{-1} (it exists), and note that f^{-1}[V] = f^{-1}[V] where LHS is the inverse-image under f and RHS is the image under 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 |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 V be this set.

    Find f^{-1} (it exists), and note that f^{-1}[V] = f^{-1}[V] where LHS is the inverse-image under f and RHS is the image under 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 g(z) be the inverse function, it means, f(g(z)) = z, therefore, g(z) + \tfrac{1}{g(z)} = z.
    What does that make g(z)=?
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let g(z) be the inverse function, it means, f(g(z)) = z, therefore, g(z) + \tfrac{1}{g(z)} = z.
    What does that make g(z)=?
    Note sure I didn't make any mistake : 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 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 f(z)=z would send circles to exactly the same circles, but if you add the \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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