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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- Jan 24th 2011, 05:45 AMrobeulerfixed 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. - Jan 24th 2011, 06:50 AMroninpro
Maybe you can use Rouche's theorem? It seems possible that

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

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

Update: This turns out to be false, for $\displaystyle z=i$. - Jan 24th 2011, 08:03 AMrobeuler
- Jan 24th 2011, 09:22 AMroninpro
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 $\displaystyle f(z)=\sin z-z$,

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

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.