দেওয়া আছে, \(r e^{i \theta} = \frac{3 + 2 i}{2 + 3 i} + \frac{1 + 5 i}{1 - 2 i}\)
প্রথম অংশ সরল করি:
\(\frac{3 + 2 i}{2 + 3 i} = \frac{(3 + 2 i)(2 - 3 i)}{(2 + 3 i)(2 - 3 i)} = \frac{6 - 9 i + 4 i - 6 i^2}{4 - 9 i^2} = \frac{6 - 5 i + 6}{4 + 9} = \frac{12 - 5 i}{13}\)
দ্বিতীয় অংশ সরল করি:
\(\frac{1 + 5 i}{1 - 2 i} = \frac{(1 + 5 i)(1 + 2 i)}{(1 - 2 i)(1 + 2 i)} = \frac{1 + 2 i + 5 i + 10 i^2}{1 - 4 i^2} = \frac{1 + 7 i - 10}{1 + 4} = \frac{-9 + 7 i}{5}\)
এখন দুটি অংশ যোগ করি:
\(r e^{i \theta} = \frac{12 - 5 i}{13} + \frac{-9 + 7 i}{5}\)
\(r e^{i \theta} = \frac{5(12 - 5 i) + 13(-9 + 7 i)}{13 \times 5}\)
\(r e^{i \theta} = \frac{60 - 25 i - 117 + 91 i}{65}\)
\(r e^{i \theta} = \frac{(60 - 117) + (-25 + 91) i}{65}\)
\(r e^{i \theta} = \frac{-57 + 66 i}{65}\)
\(r e^{i \theta} = -\frac{57}{65} + i \frac{66}{65}\)
এখানে, \(x = -\frac{57}{65}\) এবং \(y = \frac{66}{65}\)

এখন \(r\) এর মান বের করি:
\(r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{57}{65}\right)^2 + \left(\frac{66}{65}\right)^2}\)
\(r = \sqrt{\frac{3249}{4225} + \frac{4356}{4225}}\)
\(r = \sqrt{\frac{3249 + 4356}{4225}} = \sqrt{\frac{7605}{4225}}\)
\(r = \frac{\sqrt{7605}}{\sqrt{4225}} = \frac{39\sqrt{5}}{65} = \frac{3\sqrt{5}}{5} = \frac{3}{\sqrt{5}}\)

এখন \(\theta\) এর মান বের করি। যেহেতু \(x\) ঋণাত্মক এবং \(y\) ধনাত্মক, \(\theta\) দ্বিতীয় চতুর্ভাগে আছে।
\(\tan \alpha = \left|\frac{y}{x}\right| = \left|\frac{66/65}{-57/65}\right| = \left|\frac{66}{-57}\right| = \frac{66}{57}\)
\(\alpha = \tan^{-1}\left(\frac{66}{57}\right)\)
দ্বিতীয় চতুর্ভাগে \(\theta = \pi - \alpha\)
\(\theta = \pi - \tan^{-1}\left(\frac{66}{57}\right)\)

সুতরাং, \(r = \frac{3}{\sqrt{5}}\) এবং \(\theta = \pi - \tan^{-1}\left(\frac{66}{57}\right)\)।

যদি ( r e^ i theta = 3 + 2 i 2 + 3 i + 1 + 5 i 1 - 2 i ) হয়, তবে r ও θ এর মান নির্ণয় কর।