Find total number of roots inside $|z| \le 1$ for $\sin(z^{100})=z/11$. I tried to apply Rouché's theorem:
For any two complex-valued functions $f$ and $g$ holomorphic inside some region $K$ with closed contour $\partial K$ if $|g(z)| < |f(z)|$ on $\partial K$ then $f$ and $f + g$ have the same number of zeros inside $K$.
Also, $\sin(z^{100})=\sum_{n=0}^\infty \frac{(z^{100})^{2n+1}}{(2n+1)!}\cdot(-1)^n$
However, it gets me nowhere. Could you please give me a hint?
It would be a hassle to find the values of $\sin(z^{100})$ on $|z|=1$, but since $z^{100}$ and $z$ take on the same values on $|z|=1$, let's just see the modulus of $\sin(z)$ on the unit circle. The range of moduli $\sin(z)$ takes on is the same as the range of moduli $\sin(z^{100})$ takes on. (only for $|z|=1$.
We have $$\sin(x+iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$$ the modulus of which is $$\displaystyle \frac{1}{2}\sqrt{\sin^2(x)(e^{2y}+2+e^{-2y})+\cos^2(x)(e^{2y}-2+e^{-2y})}$$$$=\displaystyle \frac{1}{2}\sqrt{e^{2y}+e^{-2y}+2\sin^2(x)-2\cos^2(x)}$$ $$\displaystyle =\frac{1}{2}\sqrt{(e^{2y}-4+e^{-2y})+(4-2\cos(2x))}$$
$$\geq \displaystyle \frac{\sqrt{2}}{2}$$
On the unit circle, where $-1 \leq y \leq 1$, this is greater than $\displaystyle \frac{1}{11}$, which is the modulus of $-\displaystyle \frac{z}{11}$ on $|z|=1$.
Now, by Rouché's Theorem, since $\sin(z^{100})$ has $100$ roots inside $|z|=1$ (counting multiplicity), and $\displaystyle \left|\sin\left(z^{100}\right)\right|>\left|-\frac{z}{11}\right|$ on $|z|=1$, we have that $\displaystyle \sin(z^{100})-\frac{z}{11}$ has $100$ roots inside $|z|=1$.
Note, that since the moduli of $\sin(z^{100})$ and $\displaystyle -\frac{z}{11}$ never attain the same modulus on $|z|=1$, there are no roots on there.
