How many roots of complex equation given in an interval

Question

Given $z^4-5z+1=0$. Determine how many roots of that equation on $1 \leq |z| \leq 2$.

Do you have any idea to solve this problem? Honestly, I have tried some classic ways but that interval giving me a headache. Does that mean the roots are bounded between $1$ and $2$? Please give me your help

Answer 1

$1\leq |z|\leq 2$ means $1\leq |z|\leq 2$, i.e. that the modulus of $z$ is bounded between $1$ and $2$.
It is not difficult to prove that our polynomial has a real root very close to $\frac{1}{5}$, another real root between $\frac{3}{2}$ and $\frac{7}{4}$, no other real root. This leaves out a couple of conjugate complex roots $\zeta,\overline{\zeta}$. Since by Vieta's formulas the product of the roots equals $1$, $|\zeta|^2$ is between $\frac{20}{7}$ and $\frac{10}{3}$. It follows that $\zeta,\overline{\zeta}$ belong to the annulus $1\leq |z|\leq 2$ and our polynomial has three roots in such region.

Answer 2

You get $|5z|>|z^4+1|$ for $|z|=1$ and $|z^4|>|5z+1|$ for $|z|=2$ which allows you via Rouché to find the number of roots in the disks with radius $1$ and $2$ and per subtraction of the sets the number of roots in the annulus.