Finding the number of roots of $z^4 - 8z + 10$ in an annulus

Question

How many roots of $z^4 - 8z + 10$ lie in the annulus $1<|z|<3$?
I tried using Rouche's Theorem by looking at

$|-8z + 10| \leq 8|z| + 10 < 8\times 3 + 10 = 34$
but I can't compare this to $|z|^4$ since the lower bound of $|z|^4$ is $1$.

Answer 1

If $|z|=1$, then $|z|^4=1<|-8z+10|$. Therefore, $z^4-8z+10$ has as many zeros in $D(0,1)$ as $-8z+10$, which means that it has none.

If $|z|=3$, then $|-8z+10|<|z|^4=81$. Therefore, $z^4-8z+10$ has as many zeros in $D(0,3)$ as $z^4$, which means that it has $4$.

So, the answer is $4$.

Answer 2

Let $f(z)=z^4-8z$ and $g(z)=10$.

For $|z|=1$, then $|f|\leq9<|g|$ so in $|z|<1$ there is no root.

Since $|z^4-8z+10|\geq1$ then there is no root on $|z|=1$.

For $|z|=3$, $|f|\geq|z^4|-8|z|=57>|g|$ so in $|z|\leq3$ there are $4$ roots. This shows all $4$ roots are in annulus $1<|z|<3$.