$g(z)=z^7-2z^5+6z^3-z+1$
and choose $f(z)=2z^5-6z^3$.
On $\mid z\mid =1$, we have
$\mid f(z)-g(z) \mid=1$ and $\mid f(z) \mid=4$
so $$\mid f(z)-g(z)\mid \leq \mid f(z)\mid$$
and by Rouche's Theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk.
Now, $f(z)=2z^5-6z^3=2z^3(z^2-3)$ has 3 roots inside the unit disk, so $g(z)$ has 3 roots inside the unit disk.
Is this argument correct?