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Math Help - fixed points of sin in the complex unit disk

  1. #1
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    fixed points of sin in the complex unit disk

    I think that 0 is the only fixed point of the sin function in the complex unit disc.
    In other words sinz=z. I'm not sure how to prove it.
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  2. #2
    Senior Member roninpro's Avatar
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    Maybe you can use Rouche's theorem? It seems possible that

    |(\sin z - z)+z|< |z|

    If this is true, then you can conclude that \sin z-z and z have the same number of roots, which would be 1.

    Update: This turns out to be false, for z=i.
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  3. #3
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    Quote Originally Posted by roninpro View Post
    Maybe you can use Rouche's theorem? It seems possible that

    |(\sin z - z)+z|< |z|

    If this is true, then you can conclude that \sin z-z and z have the same number of roots, which would be 1.

    Update: This turns out to be false, for z=i.
    Hmm I see. I'm trying to think of the set of all fixed points in the unit disk now. I can't think of any others.
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  4. #4
    Senior Member roninpro's Avatar
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    You could also try to use the argument principle to measure the number of roots in the disk, but the function might be a little nasty to deal with. Putting f(z)=\sin z-z,

    \displaystyle \frac{1}{2\pi i}\oint_{\partial D(0,1)} \frac{f'(z)}{f(z)}\text{ d}z=\frac{1}{2\pi i}\oint_{\partial D(0,1)} \frac{\cos z-1}{\sin z-z}\text{ d}z

    gives the number of roots in the disk. If you can evaluate this, then you are done.

    Maybe somebody has a cleaner way of doing this.
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