Solution
$\operatorname{Re}(\mathrm{z})=\frac{\mathrm{z}+\overline{\mathrm{z}}}{2} \Rightarrow \operatorname{Re}\left(\frac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}\right)+\operatorname{Re}\left(\frac{\overline{\mathrm{z}}-1}{2 \overline{\mathrm{z}}-\mathrm{i}}\right)=2 \Rightarrow \frac{1}{2}\left[\frac{\mathrm{z}-1}{2 \mathrm{z}+\mathrm{i}}+\left(\frac{\overline{\mathrm{z}-1}}{2 \mathrm{z}+\mathrm{i}}\right)\right]+\frac{1}{2}\left[\frac{\overline{\mathrm{z}}-1}{2 \overline{\mathrm{z}}-\mathrm{i}}+\overline{\left(\frac{\overline{\mathrm{z}}-1}{2 \overline{\mathrm{z}}-\mathrm{i}}\right)}\right]=2$
\[
\begin{aligned}
& \Rightarrow \frac{z-1}{2 z+i}+\frac{\bar{z}-1}{2 \bar{z}-i}+\frac{\bar{z}-1}{2 \bar{z}-i}+\frac{z-1}{2 z+i}=4 \Rightarrow \frac{z-1}{2 z+i}+\frac{\bar{z}-1}{2 \bar{z}-i}=2 \Rightarrow(z-1)(2 \bar{z}-i)+(\bar{z}-1)(2 z+i)=2(2 z+i)(2 \bar{z}-i) \\
& \Rightarrow 2 z \bar{z}-i z-2 \bar{z}+i+2 z \bar{z}+\bar{z} i-2 z-i=8 z \bar{z}-4 z i+4 i \bar{z}+2 \Rightarrow 4 z \bar{z}+(2-3 i) z+(2+3 i) \bar{z}+2=0 \Rightarrow 4 z \bar{z}+2(z+\bar{z})-3 i(z-\bar{z})+2=0
\end{aligned}
\]
let $\mathrm{z}=\mathrm{x}+\mathrm{i}, \overline{\mathrm{z}}=\mathrm{x}-\mathrm{iy} \Rightarrow 4\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+4 \mathrm{x}+6 \mathrm{y}+2=0 \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}+\frac{3}{2} \mathrm{y}+\frac{1}{2}=0$ centre $\left(-\frac{1}{2},-\frac{3}{4}\right) \equiv(\mathrm{a}, \mathrm{b})$
radius $=\mathrm{r}=\sqrt{\frac{1}{4}+\frac{9}{16}-\frac{1}{2}}=\frac{\sqrt{5}}{4} \therefore \frac{15 \mathrm{ab}}{\mathrm{r}^{2}}=\frac{15\left(-\frac{1}{2}\right)\left(-\frac{3}{4}\right)}{\left(\frac{\sqrt{5}}{4}\right)^{2}}=\frac{45}{8} \times \frac{16}{5}=18$
.
