Example 6 In the second quadrant of the complex plane, the equation
$$
z^{6}+6 z+10=0
$$

has how many complex roots?
Let $P(z)=z^{6}+6 z+10$.
It is easy to see that $P(z)=0$ has no repeated roots.
From $P^{\prime}(z)=6\left(z^{5}+1\right)=0$, we know that when $z \in \mathbf{R}$, $P(z)_{\min }=P(-1)=5>0$.
Therefore, the equation $P(z)=0$ has no real roots.
When $z=b \mathrm{i}(b \neq 0, b \in \mathbf{R})$,
$\operatorname{Im} P(z)=6 b \neq 0$.
Thus, the equation $P(z)=0$ has no purely imaginary roots.
By Vieta's formulas, we know that the sum of the six imaginary roots of the equation $P(z)=0$ is 0, so the roots cannot all be in the left half-plane or all in the right half-plane. Since $P(z)=0$ is a real-coefficient equation, the imaginary roots appear in pairs, so the number of roots in the first quadrant is equal to the number of roots in the fourth quadrant, and the number of roots in the second quadrant is equal to the number of roots in the third quadrant.

Now we only need to discuss the number of roots of $P(z)=0$ in the first quadrant.

Let $z_{1}=r \mathrm{e}^{\mathrm{i} \theta}\left(0<\theta<\frac{\pi}{2}\right)$. Then $6 z_{1}+10$ is also in the first quadrant, and $\arg \left(6 z_{1}+10\right)<\arg z_{1}=\theta$.

Also, $z_{1}^{6}=r^{6} \mathrm{e}^{\mathrm{i} 6 \theta}=-\left(6 z_{1}+10\right)$, so $z_{1}^{6}$ is in the third quadrant.
Combining $6 \theta<3 \pi$, and $z_{1}^{6}$ is in the third quadrant, we know
$\pi<6 \theta<\frac{3 \pi}{2}$.
Thus, $6 \theta-\pi$ is the principal argument of $-z_{1}^{6}$, and it is in the interval $\left(0, \frac{\pi}{2}\right)$, i.e.,
$6 \theta-\pi=\arg \left(6 z_{1}+10\right)<\theta$.
Therefore, $0<\theta<\frac{\pi}{5}$.
Thus, the principal argument of any root $z_{1}$ of the equation $P(z)=0$ in the first quadrant $\arg z_{1} \in\left(0, \frac{\pi}{5}\right)$.

Next, we prove that the imaginary root of $P(z)=0$ in the first quadrant is unique.

Assume $z_{1} 、 z_{2}$ are two different roots of $P(z)=0$ in the first quadrant. Then
$$
\begin{aligned}
0 & =\frac{P\left(z_{1}\right)-P\left(z_{2}\right)}{z_{1}-z_{2}} \\
& =z_{1}^{5}+z_{1}^{4} z_{2}+z_{1}^{3} z_{2}^{2}+z_{1}^{2} z_{2}^{3}+z_{1} z_{2}^{4}+z_{2}^{5}+6 .
\end{aligned}
$$

Since $\arg z_{1} 、 \arg z_{2} \in\left(0, \frac{\pi}{5}\right)$, the principal argument of each term $z_{1}^{k} z_{2}^{5-k}(k=0,1, \cdots, 5)$ in equation (1) is an angle in the interval $(0, \pi)$, meaning the imaginary part of the right side of equation (1) is greater than 0, which is a contradiction. Therefore, $P(z)=0$ has only one imaginary root in the first and fourth quadrants. Thus, there are two imaginary roots in the second and third quadrants.