Example 2 Find all complex solutions of the equation $z^{11}+z=1$ with modulus 1.

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Analysis 1: Taking the modulus as a breakthrough, transform the equation to $z^{11}=1-z$. Taking the modulus on both sides and combining with $|z|=1$, we get $|z-1|=1$. Solving for $z$ and substituting back into the original equation for verification.

Solution 1: The original equation is $z^{11}=1-z$. Taking the modulus on both sides and combining with $|z|=1$, we get $|z-1|=1$. From $|z|=1$ and $|z-1|=1$, and the geometric meaning of complex numbers, it is easy to find $z=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}$. Let $w$ be a cube root of unity, then $z=-w$. Thus,
$z^{11}+z=(-w)^{11}-w=-w^{2}-w=1$,
Therefore, $z=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}$ is a solution to the original equation with modulus 1.
Analysis 2: Let $z=\cos \theta+\mathrm{i} \sin \theta, \theta \in[0, 2 \pi)$. Using the necessary and sufficient conditions for equality of complex numbers, convert the problem into a trigonometric problem to solve.

Solution 2: Let $z=\cos \theta+\operatorname{isin} \theta, \theta \in[0,2 \pi)$, substitute into the original equation, we get $\cos 11 \theta+\mathrm{i} \sin 11 \theta=1-\cos \theta-\mathrm{i} \sin \theta$.

By the necessary and sufficient conditions for equality of complex numbers, we get $\left\{\begin{array}{l}\cos 11 \theta=1-\cos \theta \\ \sin 11 \theta=-\sin \theta\end{array}\right.$
$(1)^{2}+(2)^{2}$, we get $(1-\cos \theta)^{2}+\sin ^{2} \theta=1$.
That is, $\cos \theta=\frac{1}{2}$. Since $\theta \in[0,2 \pi)$, we have $\theta=\frac{\pi}{3}$ or $\frac{5}{3} \pi$.
Clearly, $\theta=\frac{\pi}{3}$ or $\frac{5 \pi}{3}$ are solutions to the system of equations, thus $z=\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$ are solutions to the original equation with modulus 1.