AI-তৈরি নমুনা উত্তর
Given,z = 1 + 2 i 1 − 3 i z = \frac{1 + 2i}{1 - 3i} z = 1 − 3 i 1 + 2 i = ( 1 + 2 i ) ( 1 + 3 i ) ( 1 − 3 i ) ( 1 + 3 i ) = \frac{(1 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} = ( 1 − 3 i ) ( 1 + 3 i ) ( 1 + 2 i ) ( 1 + 3 i ) = 1 + 3 i + 2 i + 6 i 2 1 2 − ( 3 i ) 2 = \frac{1 + 3i + 2i + 6i^2}{1^2 - (3i)^2} = 1 2 − ( 3 i ) 2 1 + 3 i + 2 i + 6 i 2 = 1 + 5 i − 6 1 + 9 = \frac{1 + 5i - 6}{1 + 9} = 1 + 9 1 + 5 i − 6 = − 5 + 5 i 10 = \frac{-5 + 5i}{10} = 10 − 5 + 5 i = − 1 2 + 1 2 i = -\frac{1}{2} + \frac{1}{2}i = − 2 1 + 2 1 i
Now, modulus r r r r = ∣ z ∣ = ( − 1 2 ) 2 + ( 1 2 ) 2 = 1 4 + 1 4 = 1 2 = 1 2 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}} r = ∣ z ∣ = ( − 2 1 ) 2 + ( 2 1 ) 2 = 4 1 + 4 1 = 2 1 = 2 1
Argument θ \theta θ z z z θ = π − tan − 1 ∣ 1 / 2 − 1 / 2 ∣ = π − tan − 1 ( 1 ) = π − π 4 = 3 π 4 \theta = \pi - \tan^{-1}\left|\frac{1/2}{-1/2}\right| = \pi - \tan^{-1}(1) = \pi - \frac{\pi}{4} = \frac{3\pi}{4} θ = π − tan − 1 − 1/2 1/2 = π − tan − 1 ( 1 ) = π − 4 π = 4 3 π
Therefore, the polar form is:1 2 ( cos 3 π 4 + i sin 3 π 4 ) \frac{1}{\sqrt{2}}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) 2 1 ( cos 4 3 π + i sin 4 3 π )