Show $2z^5-6z^2+z+1$ has three zeroes in $\{z : 1<\vert z \vert < 2\}$

Question

I need to show

$$2z^5-6z^2+z+1$$

has three zeroes on $\{z : 1< \vert z \vert < 2\}.$

So do I split in into cases for when $\vert z \vert= 1$

and when $\vert z \vert = 2$ by Rouches theorem I can take one function to be $2z^5+z+1$ and the other be $h(z) = -6z^2$

Am I doing ok so far?

Answer 1

When $|z|=1$, you take $f_1(z)=-6z^2$ and $g_1(z)=2z^5+z+1$. So, if $|z|=1$,\begin{align}|g_1(z)|&\leqslant4\\&<6\\&=|f_1(z)|.\end{align}Therefore, your function has $2$ zeros (counted with their multiplicities) when $|z|<1$ (and no zero when $|z|=1$).

When $|z|=2$, you take $f_2(z)=2z^5$ and $g_2(z)=-6z^2+z+1$. So, if $|z|=2$,\begin{align}|g_1(z)|&\leqslant27\\&<64\\&=|f_1(z)|.\end{align}Therefore, your function has $5$ zeros (counted with their multiplicities) when $|z|<2$.

The conclusion is that it has three zeros when $1<|z|<2$.

Answer 2

Use as $g(z)=-6z^{2}+z,$ $ f(z)=2z^{5}+1$ and| $-6z^{2}+z|\leq 6|z|^{2}+|z|\leq |2z^{5}|+1=2|z|^{5}+1$ on |z|=2. Thus f and f+g have the same number of zeros inside |z|<2 that is five. Then use as $g(z)=2z^{5}+z+1, $ and $f(z)=-6z^{2}$ and in this case ,$|2z^{5}+z+1|\leq 4\leq |-6z^{2}|=6$ on $|z|=1$. So the initial function has as many roots as $-6z^{2}$ that is two. Therefore we get exactly three roots in $1<|z|<2$ !!!