Question 116, Given the complex number $\mathrm{z}=\mathrm{t}+3+\sqrt{3} \mathrm{i}$, where $\mathrm{t}$ is a complex number such that $\frac{\mathrm{t}+3}{\mathrm{t}-3}$ is a pure imaginary number.
(1) Find the equation of the trajectory of point $\mathrm{P}$ corresponding to $\mathrm{z}$ in the complex plane;
(2) Find the range of $\arg z$;
(3) Find the maximum and minimum values of $|z-1|^{2}+|z+1|^{2}$.
Question 116, Solution: Let $\frac{t+3}{t-3}=k i(k \in R \backslash\{0\})$, then $t=\frac{3+3 k i}{k i-1}=\frac{3\left(k^{2}-1\right)-6 k i}{k^{2}+1}$. Therefore, $z=\frac{3\left(k^{2}-1\right)-6 k i}{k^{2}+1}+3+\sqrt{3} i=\frac{6 k^{2}+\left(\sqrt{3}\left(k^{2}+1\right)-6 k\right) i}{k^{2}+1}$
(1) Let the complex number corresponding to point $P$ be $x+y i$, then $\left\{\begin{array}{c}x=\frac{6 k^{2}}{k^{2}+1} \\ y=\frac{\sqrt{3}\left(k^{2}+1\right)-6 k}{k^{2}+1}=\sqrt{3}-\frac{6 k}{k^{2}+1}\end{array}\right.$
Clearly, $x \in(0,6)$, and $k^{2}=\frac{x}{6-x}$, and we have
$$
\begin{array}{l}
(\sqrt{3}-y)^{2}=\frac{36 k^{2}}{\left(k^{2}+1\right)^{2}}=\frac{36 \cdot \frac{x}{6-x}}{\left(\frac{x}{6-x}+1\right)^{2}}=x(6-x) \\
\Rightarrow(x-3)^{2}+(y-\sqrt{3})^{2}=9
\end{array}
$$

Thus, the trajectory equation of point $P$ is $(x-3)^{2}+(y-\sqrt{3})^{2}=9$ excluding the points $(0, \sqrt{3})$ and $(6, \sqrt{3})$. As shown in the figure below.
(2) As shown in the figure, draw $O A$ tangent to $\odot Q$ at $A$, then $\angle Q A O=\arcsin \frac{\sqrt{3}}{2}=\frac{\pi}{3}$, and the angle between ray $O Q$ and the positive x-axis is $\frac{\pi}{6}$. Therefore, the range of $\arg z$ is $\left[0, \frac{\pi}{2}\right) \cup\left[\frac{11 \pi}{6}, 2 \pi\right)$.
(3) Since $|z-1|^{2}+|z+1|^{2}=(z-1)(\bar{z}-1)+(z+1)(\bar{z}+1)=2\left(|z|^{2}+1\right)$, we only need to find the maximum and minimum values of $|z|$.
For any point $X$ on $\odot Q$, we have $|O Q|-|Q X| \leq|O X| \leq|O Q|+|Q X|$, so the range of $|O X|$ is $[2 \sqrt{3}-3,2 \sqrt{3}+3]$, which means the maximum and minimum values of $|z|$ are $2 \sqrt{3}+3$ and $2 \sqrt{3}-3$ respectively. Therefore, the maximum value of $|z-1|^{2}+|z+1|^{2}$ is $2\left((2 \sqrt{3}+3)^{2}+1\right)=4(11+6 \sqrt{3})$, and the minimum value is $2\left((2 \sqrt{3}-3)^{2}+1\right)=4(11-6 \sqrt{3})$.