A) \[\frac{3\pi }{4}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{5\pi }{6}\]
D) \[-\frac{3\pi }{4}\]
Correct Answer: B
Solution :
\[\operatorname{z}=\frac{-\,2(1+2i)}{3+i}\] \[=\,\,\frac{-\,2-4i}{3+i}=\frac{-2-4i}{3+i}\times \frac{3-i}{3-i}\] \[=\,\,\frac{-\,6+2i-12i+4{{i}^{2}}}{10}\] \[=\,\,\frac{-6-10i-4}{10}=\frac{-10-10i}{10}=-1-i\] \[\operatorname{z} = - 1- i = r \left( cos \theta + i sin \theta \right)\] On comparing real and imaginary part on both sides, we get \[\operatorname{r}\,cos\,\,\theta =-1\] ... (i) \[\operatorname{r}\,\sin \,\,\theta =-1\] ... (ii) On dividing eq. (ii) by (i), we get \[\frac{\operatorname{r}\,\sin \,\,\theta }{\operatorname{r}\,cos\,\,\theta }=\frac{-1}{-1}\] \[\tan \theta =1=\tan \frac{\pi }{4}\] \[\Rightarrow \,\,\,\,\theta =\frac{\pi }{4}\,\,\] \[\therefore \,\,\,\,\theta =\frac{\pi }{4}\,\,\]
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