যদি  $r e^{i \theta} = \frac{3 + 2 i}{2 + 3 i} + \frac{1 + 5 i}{1 - 2 i}$ হয়, তবে r ও θ এর মান নির্ণয় কর।

দেয়া আছে, $r \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{3+2 \mathrm{i}}{2+3 \mathrm{i}}+\frac{1+5 \mathrm{i}}{1-2 \mathrm{i}}$

$\begin{array}{l} \Rightarrow \mathrm{r} \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{(3+2 \mathrm{i})(2-3 \mathrm{i})}{(2+3 \mathrm{i})(2-3 \mathrm{i})}+\frac{(1+5 \mathrm{i})(1+2 \mathrm{i})}{(1-2 \mathrm{i})(1+2 \mathrm{i})} \\ \Rightarrow \mathrm{r} \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{6-9 \mathrm{i}+4 \mathrm{i}-6 \mathrm{i}^{2}}{4-9 \mathrm{i}^{2}}+\frac{1+2 \mathrm{i}+5 \mathrm{i}+10 \mathrm{i}^{2}}{1-4 \mathrm{i}^{2}} \\ \Rightarrow \mathrm{r} \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{(6+6)+\mathrm{i}(4-9)}{4+9}+\frac{(1-10)+\mathrm{i}(2+5)}{1+4}\left[\because \mathrm{i}^{2}=-1\right] \\ \Rightarrow \mathrm{r} \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{12-5 \mathrm{i}}{13}+\frac{-9+7 \mathrm{i}}{5} \\ \Rightarrow \mathrm{r} \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{12}{13}-\frac{5}{13} \mathrm{i}-\frac{9}{5}+\frac{7}{5} \mathrm{i} \\ \Rightarrow \mathrm{r} \cdot \mathrm{e}^{\mathrm{i} \theta}=\frac{-57}{65}+\frac{66}{65} \mathrm{i} \\ \therefore \mathrm{r}=\sqrt{\left(\frac{-57}{65}\right)^{2}+\left(\frac{66}{65}\right)^{2}}=\frac{\sqrt{32+19+4356}}{65}=\frac{\sqrt{7605}}{65}=\frac{39 \sqrt{5}}{65}=\frac{3}{\sqrt{5}} \text { (Ans.) } \\ \theta=\pi-\tan ^{-1}\left(\frac{\frac{66}{65}}{\frac{57}{65}}\right)=\pi-\tan ^{-1}\left(\frac{66}{57}\right) \text { (Ans.) } \end{array}$