Example 1 (Question from the 9th "Hope Cup" Invitational Competition) Given a complex number $z$ satisfying $|z|=1$, then the range of the function $\mid z^{2}+\mathrm{i} z^{2}+1 \mid$ is ( ).
A. $\left[1-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}+1\right]$
B. $[\sqrt{2}-1, \sqrt{2}+1]$
C. $[2 \sqrt{2}-1,2 \sqrt{2}+1]$
D. $[2-\sqrt{2}, \sqrt{2}+1]$
Solution: Choose B. Reason: It is easy to know that $\left|\mathrm{i} z^{2}+z^{2}+1\right|=\left|(1+\mathrm{i}) z^{2}+1\right|=|1+\mathrm{i}| \cdot\left|z^{2}+\frac{1}{1+\mathrm{i}}\right|=$ $\sqrt{2}\left|z^{2}+\frac{1-\mathrm{i}}{2}\right|$.
Let $w=z^{2}$, then by $|z|=1$, we know $|w|=1$.
The geometric meaning of the function $f(w)=\left|w-\left(\frac{-1+\mathrm{i}}{2}\right)\right|$ is the distance from a point on the unit circle of the complex plane $w$ to a fixed point $-\frac{1}{2}+\frac{1}{2} \mathrm{i}$, as shown in Figure $9-1$. $P$ is a point on the unit circle $|w|=1$, $A\left(-\frac{1}{2}, \frac{1}{2}\right)=$ $-\frac{1}{2}+\frac{1}{2} \mathrm{i}, P_{1} P_{2}$ is the diameter passing through the origin $O$ and $A$. Thus,
$$
\begin{array}{l}
\min _{|w|=1} f(w)=f\left(P_{1}\right)=\left|P_{1} A\right|=1-\frac{\sqrt{2}}{2}, \\
\min _{|w|=1} f(w)=f\left(P_{2}\right)=\left|P_{2} A\right|=1+\frac{\sqrt{2}}{2} .
\end{array}
$$

Therefore, the range of the function $\left|\mathrm{i} z^{2}+z^{2}+1\right|$ is
$$
\left[\sqrt{2}\left(1-\frac{\sqrt{2}}{2}\right), \sqrt{2}\left(1+\frac{\sqrt{2}}{2}\right)\right]=[\sqrt{2}-1, \sqrt{2}+1] \text {. }
$$

Alternative solution: Let $z=\cos \theta+\mathrm{i} \sin \theta(0 \leqslant \theta \leqslant 2 \pi)$, thus $z^{2}=\cos 2 \theta+\mathrm{i} \sin 2 \theta$, then
$$
\begin{array}{l}
\left|\mathrm{i} z^{2}+z^{2}+1\right|=|\mathrm{i} \cos 2 \theta-\sin 2 \theta+\cos 2 \theta+\mathrm{i} \sin 2 \theta+1| \\
=|(\cos 2 \theta-\sin 2 \theta+1)+\mathrm{i}(\cos 2 \theta+\sin 2 \theta)| \\
=\sqrt{(\cos 2 \theta-\sin 2 \theta+1)^{2}+(\cos 2 \theta+\sin 2 \theta)^{2}} \\
=\sqrt{3+2(\cos 2 \theta-\sin 2 \theta)} \\
=\sqrt{3+2 \sqrt{2} \sin \left(45^{\circ}-2 \theta\right)}(0 \leqslant \theta \leqslant 2 \pi)
\end{array}
$$

Thus, $\quad\left|\mathrm{i} z^{2}+z^{2}+1\right|_{\max }=\sqrt{3+2 \sqrt{2}}=1+\sqrt{2}$,
$$
\left|i z^{2}+z^{2}+1\right|_{\text {min }}=\sqrt{3-2 \sqrt{2}}=\sqrt{2}-1 .
$$

Therefore, the required range is $[\sqrt{2}-1, \sqrt{2}+1]$.