Example 6. Find the number of zeros of the function

$$
F(z)=z^{8}-4 z^{5}+z^{2}-1
$$

inside the unit circle $|z|<1$.
Solution. Let us represent the function $F(z)$ as the sum of two functions $f(z)$ and $\varphi(z)$, which we choose, for example, as follows:

$$
f(z)=-4 z^{5}, \quad \varphi(z)=z^{8}+z^{2}-1
$$

Then on the circle $|z|=1$ we will have

$$
\begin{aligned}
& |f(z)|=\left|-4 z^{5}\right|=4 \\
& |\varphi(z)|=\left|z^{8}+z^{2}-1\right| \leqslant\left|z^{8}\right|+\left|z^{2}\right|+1=3
\end{aligned}
$$

Thus, on the boundary $|z|=1$ of the circle, the inequality $|f(z)|>|\varphi(z)|$ holds. The function $f(z)=-4 z^{5}$ has a fivefold zero at the origin. By Rouché's theorem, the function

$$
F(z)=f(z)+\varphi(z)=z^{8}-4 z^{5}+z^{2}-1
$$

has five zeros in the circle $|z|<1$. Note that a different choice of functions $f(z)$ and $\varphi(z)$ is also possible, for example, such as:

$$
f(z)=z^{8}-4 z^{5}, \quad \varphi(z)=z^{2}-1
$$