Example 6 Find all real number pairs $\left(\lambda_{1}, \lambda_{2}\right)$, such that the equation $\left(x+\mathrm{i} \lambda_{1}\right)^{n}+\left(x+\mathrm{i} \lambda_{2}\right)^{n}=0$
has all real solutions, and find the real solutions of the equation.
Let $x$ be a complex solution of (1). From (1), we have $\left(x+\mathrm{i} \lambda_{1}\right)^{n}=-\left(x+\mathrm{i} \lambda_{2}\right)^{n}$,
so $\left|x+\mathrm{i} \lambda_{1}\right|^{n}=\left|x+\mathrm{i} \lambda_{2}\right|^{n}$,
so $\left|x+i \lambda_{1}\right|=\left|x+i \lambda_{2}\right|$.
In the complex plane, let the points corresponding to $-i \lambda_{1},-i \lambda_{2}$ be $A, B$, respectively. Then the point corresponding to $x$ lies on the perpendicular bisector of segment $A B$. For $x$ to be a real number, the perpendicular bisector of $A B$ must be the $x$-axis,
so $\lambda_{1}+\lambda_{2}=0$. Let $\lambda_{1}=-\lambda_{2}=\lambda \in \mathbf{R}$, substituting into (1) gives $(x+\mathrm{i} \lambda)^{n}=-(x-\mathrm{i} \lambda)^{n}$
If $\lambda=0$, then $x=0$ is the only real solution. If $\lambda \neq 0$, from (3) we get $\left(\frac{x+\mathrm{i} \lambda}{x-\mathrm{i} \lambda}\right)^{n}=-1$. So $\frac{x+\mathrm{i} \lambda}{x-\mathrm{i} \lambda}=u_{k}$.
Here
$$
\begin{aligned}
u_{k} & =\cos \left(\frac{\pi}{n}+\frac{2 k \pi}{n}\right)+\mathrm{i} \sin \left(\frac{\pi}{n}+\frac{2 k \pi}{n}\right) \\
& =\cos \frac{(2 k+1) \pi}{n}+\mathrm{i} \sin \frac{(2 k+1) \pi}{n}, k=0,1,2, \cdots, n-1 .
\end{aligned}
$$

So $x=\mathrm{i} \lambda \frac{u_{k}+1}{u_{k}-1}$, i.e., $x=\lambda \cot \frac{(2 k+1) \pi}{2 n}, k=0,1, \cdots, n-1$.
Therefore, the $n$ real solutions of the equation are $x_{k}=\lambda \cot \frac{(2 k+1) \pi}{2 n}, k=0,1,2, \cdots, n-1$.