Question 112, Given complex numbers $z_{i} (i=1, 2, 3, 4, 5)$ satisfy: $\left\{\begin{array}{c}\left|z_{1}\right| \leq 1,\left|z_{2}\right| \leq 1 \\ \left|2 z_{3}-\left(z_{1}+z_{2}\right)\right| \leq\left|z_{1}-z_{2}\right| \\ \left|2 z_{4}-\left(z_{1}+z_{2}\right)\right| \leq\left|z_{1}-z_{2}\right| \\ \left|2 z_{5}-\left(z_{3}+z_{4}\right)\right| \leq\left|z_{3}-z_{4}\right|\end{array}\right.$ Try to find the maximum value of $\left|z_{5}\right|$.
Question 112, Solution: Establish a complex plane, and let $\mathrm{Z}_{\mathrm{i}}$ represent the point corresponding to $\mathrm{Z}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 5)$. According to the conditions, $Z_{1}$ and $Z_{2}$ are inside the unit circle $\odot 0$ centered at the origin 0, $\mathrm{Z}_{3}$ and $\mathrm{Z}_{4}$ are inside the circle $\odot \mathrm{P}$ with $\mathrm{Z}_{1} \mathrm{Z}_{2}$ as its diameter, and $\mathrm{Z}_{5}$ is inside the circle $\odot P$ with $\mathrm{Z}_{3} \mathrm{Z}_{4}$ as its diameter.

Let the radius of $\odot P$ be $r_{1}$, and the radius of $\odot Q$ be $r_{2}$. Then, according to the power of a point theorem, we have $O P \leq \sqrt{1-\mathrm{PZ}_{1} \cdot \mathrm{PZ}_{2}} \leq \sqrt{1-\mathrm{r}_{1}{ }^{2}}$, $\mathrm{PQ} \leq \sqrt{\mathrm{r}^{2}-\mathrm{QZ} \cdot Q Z_{4}} \leq \sqrt{\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}$, and $Q \mathrm{Z}_{5} \leq \mathrm{r}_{2}$. Therefore, we know
$$
\begin{array}{l}
\left|z_{5}\right| \leq O P+P Q+Q Z_{5} \leq \sqrt{1-r_{1}{ }^{2}}+\sqrt{r_{1}{ }^{2}-r_{2}{ }^{2}}+r_{2} \\
\leq \sqrt{3\left[\left(1-r_{1}{ }^{2}\right)+\left(r_{1}{ }^{2}-r^{2}{ }^{2}\right)+r_{2}{ }^{2}\right]}=\sqrt{3}
\end{array}
$$

On the other hand, if we take $z_{1}=\frac{\sqrt{2}+1}{\sqrt{3}} ; z_{2}=\frac{-\sqrt{2}+i}{\sqrt{3}} ; z_{3}=\frac{1+2 i}{\sqrt{3}} ; z_{4}=\frac{-1+2 i}{\sqrt{3}} ; z_{5}=\sqrt{3} \mathrm{i}$, then the conditions are satisfied.