I need to prove that the number of zeros of $z^4-z^3-4z+1$ in the ring $\{ 1 < |z| < 2 \}$ is equal to $3$.
What I have done so far; I've proved there is only one zero in the $\{|z|\leq1\}$. So the only thing remaining is to prove there is no zeros in the $\{|z|\geq2\}$. This part, however, creates a lot of issues for me. I've tried the classic way using Rouche's theorem and even adding new polynomials in order to use it, but nothing helped.
Also tried assuming $z^4-z^3-4z+1=0$ and $|z|\geq2$ with some clever use of inequalites, that didn't help either. I don't know if I'm missing something, but according to WA, one zero is $\approx1.9325$ so that might be causing troubles in my opinion.
It can be proved by Rouche's theorem.
Hint: $$|z^4 + 1| > |-z^3 - 4z|$$ on $|z| = 2$.
You can't prove this inequality by just using triangle inequality.
One way to prove the inequality is by letting $z = 2e^{i\theta}$ then the problem is equivalent with $$257 + 32 \cos 4\theta > 256 \cos^2 \theta$$
You can prove that this is indeed true for all $\theta\in\mathbb{R}$