If ${\mathrm{z}}_1 = 1 - 2\mathrm{i} ; {\mathrm{z}}_2 = 1+\mathrm{i}$ and ${\mathrm{z}}_3 = 3 + 4\mathrm{i}$ then $\left(\frac{1}{{\mathrm{z}}_1}+ \frac{3}{{\mathrm{z}}_2}\right)\frac{{\mathrm{z}}_3}{{\mathrm{z}}_2} =$

$$\begin{gathered} {\mathrm{z}}_1=1-2\mathrm{i} {\mathrm{z}}_2=1+\mathrm{i} {\mathrm{z}}_3=3+4\mathrm{i} \\ \Rightarrow \frac{1}{{\mathrm{z}}_1}=\frac{1}{1-2_{\mathrm{i}}}=\frac{1+2\mathrm{i}}{(1-2\mathrm{i})(1+2\mathrm{i})}=\frac{1+2\mathrm{i}}{1+4}=\frac{1+2\mathrm{i}}{5} \\ \mathrm{and} \frac{3}{{\mathrm{z}}_2}=\frac{3}{1+\mathrm{i}}=\frac{3(1-\mathrm{i})}{(1+\mathrm{i})(1-\mathrm{i})}=\frac{3(1-\mathrm{i})}{1+1}=\frac{3(1-\mathrm{i})}{2}, \\ \frac{z_3}{{\mathrm{z}}_2}=\frac{3+4\mathrm{i}}{1+\mathrm{i}}=\frac{(3+4\mathrm{i}) (1-\mathrm{i})}{(1+\mathrm{i}) (1-\mathrm{i})}=\frac{(3-4{\mathrm{i}}^2)-3\mathrm{i}+4\mathrm{i}}{1+1} \\ \Rightarrow \frac{{\mathrm{z}}_3}{{\mathrm{z}}_2}=\frac{7+\mathrm{i}}{2} \\ \therefore \left(\frac{1}{{\mathrm{z}}_1}+\frac{3}{{\mathrm{z}}_2}\right)\frac{{\mathrm{z}}_3}{{\mathrm{z}}_2}=\left(\frac{1+2\mathrm{i}}{5}+\frac{3(1-\mathrm{i})}{2}\right) \left(\frac{7+\mathrm{i}}{2}\right) \\ =\left(\frac{2+4\mathrm{i}+15-15\mathrm{i}}{10}\right) \left(\frac{7+\mathrm{i}}{2}\right) \\ =\frac{(17-11\mathrm{i}) (7+\mathrm{i})}{20} \\ =\frac{119-77\mathrm{i}+17\mathrm{i}-11{\mathrm{i}}^2}{20} \\ =\frac{130-60\mathrm{i}}{20} \\ =\frac{13}{2}-3\mathrm{i} \end{gathered}$$