Thread: 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.

Maybe you can use Rouche's theorem? It seems possible that



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

Update: This turns out to be false, for .

Originally Posted by roninpro

Maybe you can use Rouche's theorem? It seems possible that



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

Update: This turns out to be false, for .

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.

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 ,



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.