Example 4. Find the number of roots in the right half-plane $\operatorname{Re} z>0$ of the equation

$$
Q_{5}(z) \equiv z^{5}+z^{4}+2 z^{3}-8 z-1=0
$$
Solution. By the argument principle, the number of zeros inside the contour $C$ is

$$
N=\frac{1}{2 \pi} \Delta_{C} \operatorname{Arg} Q_{5}(z)
$$

where the contour $C$ consists of the semicircle $C_{R}:|z|=R, \operatorname{Re} z>0$, and its diameter on the imaginary axis; the radius $R$ is taken to be so large that all the zeros of the polynomial $Q_{5}(z)$, located in the right half-plane, fall inside the semicircle $|z|>0$. We have

$$
Q_{5}(z)=z^{5}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
$$

Hence,

$$
\operatorname{Arg} Q_{5}(z)=\operatorname{Arg}\left[z^{3}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)\right]=
$$

$$
\begin{aligned}
& =\operatorname{Arg} z^{5}+\operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)= \\
& =5 \operatorname{Arg} z+\operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
\end{aligned}
$$

The increment of the argument of $Q_{5}(z)$ when traversing the semicircle $C_{R}$ in the positive direction will be

$$
\Delta_{C_{R}} \operatorname{Arg} Q_{5}(z)=5 \Delta_{C_{R}} \operatorname{Arg} z+\Delta_{C_{R}} \operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
$$

In this equality, we take the limit as $R \rightarrow \infty$:

$$
\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} Q_{5}(z)=5 \lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} z+\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)
$$

Both limits on the right-hand side exist and are equal to

$$
\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} z=\pi, \quad \lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg}\left(1+\frac{1}{z}+\frac{2}{z^{2}}-\frac{8}{z^{4}}-\frac{1}{z^{5}}\right)=0
$$

Thus,

$$
\lim _{R \rightarrow \infty} \Delta_{C_{R}} \operatorname{Arg} Q_{5}(z)=5 \pi
$$

Now let the point $z$ move along the imaginary axis from $z=i R$ to $z=-i R$. Let $z=i t, -R \leqslant t \leqslant R$. Then

$$
\dot{Q}_{5}(i t)=u(t)+i v(t)=t^{4}-1+i\left(t^{5}-2 t^{3}-8 t\right)
$$

from which

$$
\left\{\begin{array}{l}
u=t^{4}-1 \\
v=t^{5}-2 t^{3}-8 t
\end{array}\right.
$$

These are the parametric equations of the curve described by the point $w=Q_{5}(z)$ in the $(\boldsymbol{u}, \boldsymbol{v})$ plane as the point $z$ traverses the imaginary axis from top to bottom. To construct this curve, we find the points of its intersection with the coordinate axes $O u$ and $O v$. Setting $u$ and $v$ to zero, we get respectively

$$
\begin{aligned}
& t^{4}-1=0, \quad \text { or } t= \pm 1 \\
& t^{5}-2 t^{3}-8 t=0, \text { or } t= \pm 2, \quad t=0
\end{aligned}
$$

Note that equations (2) and (3) do not have common real roots, so the polynomial $Q_{5}(z)$ has no zeros on the imaginary axis. Therefore, the application of the argument principle to the contour is valid. The roots of equations (2) and (3) are arranged in descending order, i.e., in the order of traversal of the contour, and we find the corresponding values of $\boldsymbol{u}$ and $v$:

| № | $t$ | $u$ | $v$ |
| :---: | ---: | ---: | ---: |
| 1 | 2 | 15 | 0 |
| 2 | 1 | 0 | -9 |
| 3 | 0 | -1 | 0 |
| 4 | -1 | 0 | 9 |
| 5 | -2 | 15 | 0 |

![](https://cdn.mathpix.com/cropped/2024_05_22_f7d63c3a5e94c3f2f1bbg-113.jpg?height=482&width=666&top_left_y=219&top_left_x=137)

Figure 12
Further,

$$
\begin{aligned}
& \lim _{t \rightarrow \pm \infty} u=+\infty \\
& \lim _{t \rightarrow \pm \infty} v= \pm \infty
\end{aligned}
$$

These data allow us to construct the curve of interest (Figure 12).

From Figure 12, it is clear that the vector $w=Q_{5}(z)$ will rotate by an angle $\varphi=3 \pi$ in the negative direction. Therefore,

$\Delta_{C} \operatorname{Arg}_{5}(z)=5 \pi-3 \pi=2 \pi$,

from which the number of zeros in the right half-plane will be

$$
N=\frac{2 \pi}{2 \pi}=1
$$