Solution:

$$Let, \mathrm{z}=\frac{\overline{1}+3 \sqrt{3} \mathrm{i}}{\sqrt{3}+2 \mathrm{i}}\\$$ $$\begin{aligned} =& \frac{1+3 \sqrt{3} \mathrm{i}}{\sqrt{3}+2 \mathrm{i}} \times \frac{\sqrt{3}-2 \mathrm{i}}{\sqrt{3}-2 \mathrm{i}}\\ =& \frac{\sqrt{3}-2 \mathrm{i}+9 \mathrm{i}-6 \sqrt{3} \mathrm{i}^{2}}{(\sqrt{3})^{2}-(2 \mathrm{i})^{2}} \\ =& \frac{\sqrt{3}+7 \mathrm{i}+6 \sqrt{3}}{3+4} \\ =& \frac{7 \sqrt{3}+7 \mathrm{i}}{7} \\ =& \frac{7(\sqrt{3}+\mathrm{i})}{7} \\ \mathrm{z}=& \sqrt{3}+\mathrm{i}=\mathrm{a}+\mathrm{ib} \text { form } \end{aligned}$$ $Here, \mathrm{a}=\sqrt{3} \& \mathrm{~b}=1\\ $ …