Roots of $z \sin(z) =1$

Question

this question has been answered here ,but I still have many questions. So I applied Rouche to a circle of radius $(n+\frac{1}{2}) \pi$, and set $f(z)=z \sin(z) -1$ And $g(z)=z \sin(z)$. Then I obtained the inequality $|\sin(z)|>\frac{2}{\pi}$. So by Rouche, $f$ and $g$ has the same number of roots on the disk.

$g$ has $2n+1$ roots on the disk, all of which should be on the real line portion of the disk. However, by graphing $f$, I think I see $2n+2$ roots on real line portion of the disk. here is the graph

Can anyone explain where I made the mistake? Thanks in advance.

Answer 1

You are correct, within the interval $[-r_n,r_n]$, $r_n=(n+\frac12)\pi$ the function $g(z)=z\sin(z)$ has $2n+2$ roots if counted with multiplicity, as there is a double root at $z=0$. Since this number, including multiplicity, is preserved under perturbations as characterized in the Rouché theorem, also the function $f(z)=z\sin(z)-1$ has $2n+2$ roots in the corresponding region.