# fixed points of sin in the complex unit disk

• Jan 24th 2011, 05:45 AM
robeuler
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.
• Jan 24th 2011, 06:50 AM
roninpro
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 $z=i$.
• Jan 24th 2011, 08:03 AM
robeuler
Quote:

Originally Posted by roninpro
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.
• Jan 24th 2011, 09:22 AM
roninpro
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.