Question 117: Let complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2}$ satisfy $\operatorname{Re}\left(\mathrm{z}_{1}\right)>0, \operatorname{Re}\left(\mathrm{z}_{2}\right)>0$, and $\operatorname{Re}\left(\mathrm{z}_{1}^{2}\right)=\operatorname{Re}\left(\mathrm{z}_{2}^{2}\right)=2$ (where $\operatorname{Re}(z)$ denotes the real part of the complex number $z$).
(1) Find the minimum value of $\operatorname{Re}\left(z_{1} z_{2}\right)$;
(2) Find the minimum value of $\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|\overline{z_{1}}-z_{2}\right|$.
Question 117, Solution: Let $z_{1}=a_{1}+b_{1} i, z_{2}=a_{2}+b_{2} i$, then $a_{1}>0, b_{1}>0, a_{1}^{2}-b_{1}^{2}=$ $a_{2}^{2}-b_{2}^{2}=2$.
(1) According to the conditions and the Cauchy-Schwarz inequality, we have:
$$
\begin{array}{l}
\operatorname{Re}\left(z_{1} z_{2}\right)=a_{1} a_{2}-b_{1} b_{2}=\sqrt{\left(2+b_{1}^{2}\right)\left(2+b_{2}^{2}\right)}-b_{1} b_{2} \\
\geq 2+\left|b_{1} b_{2}\right|-b_{1} b_{2} \geq 2
\end{array}
$$

Equality holds when $z_{1}=z_{2}=\sqrt{2}$.
(2) Note that $\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|\overline{z_{1}}-z_{2}\right|=\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|z_{1}-\overline{z_{2}}\right|$.
Establish a Cartesian coordinate system, set $Z_{1}(a_{1}, b_{1}), Z_{2}(a_{2}, -b_{2})$, then $Z_{1}, Z_{2}$ are both on the hyperbola $C: \frac{x^{2}}{2}-$ $\frac{y^{2}}{2}=1$, with left and right foci at $F_{1}(-2,0), F_{2}(2,0)$, and left and right directrices at $l_{1,2} = \pm 1$, and eccentricity $e=\sqrt{2}$. Let $d(A, B)$ be the distance between points $A$ and $B$, and $d(A, 1)$ be the distance from point $A$ to line 1. Then we know:
$$
\begin{array}{l}
\left|z_{1}+2\right|+\left|\overline{z_{2}}+2\right|-\left|z_{1}-\overline{z_{2}}\right| \\
=d\left(Z_{1}, F_{1}\right)+d\left(Z_{2}, F_{1}\right)-d\left(Z_{1}, Z_{2}\right) \\
\geq d\left(Z_{1}, F_{1}\right)+d\left(Z_{2}, F_{1}\right)-\left(d\left(Z_{1}, F_{2}\right)+d\left(Z_{2}, F_{2}\right)\right) \\
=e\left[d\left(Z_{1}, l_{1}\right)+d\left(Z_{2}, l_{1}\right)-\left(d\left(Z_{1}, l_{2}\right)+d\left(Z_{2}, l_{2}\right)\right)\right] \\
=\sqrt{2}\left[\left(d\left(Z_{1}, l_{1}\right)-d\left(Z_{1}, l_{2}\right)\right)+\left(d\left(Z_{2}, l_{1}\right)-d\left(Z_{2}, l_{2}\right)\right)\right] \\
=\sqrt{2}(2+2)=4 \sqrt{2}
\end{array}
$$

Equality holds when $z_{1}=z_{2}=2+\sqrt{2} i$.