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AI-তৈরি নমুনা উত্তর

Given,
z = \frac{1 + 2i}{1 - 3i}
= \frac{(1 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)}
= \frac{1 + 3i + 2i + 6i^2}{1^2 - (3i)^2}
= \frac{1 + 5i - 6}{1 + 9}
= \frac{-5 + 5i}{10}
= -\frac{1}{2} + \frac{1}{2}i

Now, modulus r:
r = |z| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

Argument \theta (since z lies in the second quadrant):
\theta = \pi - \tan^{-1}\left|\frac{1/2}{-1/2}\right| = \pi - \tan^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}

Therefore, the polar form is:
\frac{1}{\sqrt{2}}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)