Find number of elements $z$, $1<|z|<2$ satisfying $f(z)=0$ where $f(z)=z^5+z^3+5z^2+2$.

Question

Let $g(z)=z^5$ then

$\forall z$ s.t $|z|=2$

$|g(z)-f(z)|\leq |2|^3+5|2|^2+2< |2|^5=|g(z)|$

Therefore, By Rouché's $f(z) $ has 5 zeroes in $D(0,2)$. I do not know how to proceed any further. Any Solutions?

Answer 1

Now, apply Rouché's theorem on $D(0,1)$, with $g(z)=5z^2$. Then$$\lvert z\rvert=1\implies\bigl\lvert g(z)-f(z)\bigr\rvert\leqslant4<5=\bigl\lvert g(z)\bigr\rvert.$$So, there are $2$ zeros on $D(0,1)$. Therefore, there are $3$ zeros in the region that concerns you.