A) \[45{}^\circ \]
B) \[135{}^\circ \]
C) \[225{}^\circ \]
D) \[240{}^\circ \]
Correct Answer: C
Solution :
\[\left( \frac{-1-3\,i}{2+i} \right)=\frac{-1-3i}{2+i}\times \frac{2-i}{2-i}\] \[=\frac{-2+i-6i+3{{i}^{2}}}{4+1}\] \[=\frac{-2-5i-3}{5}=\frac{-5-5i}{5}=-1-i\] \[\therefore \]Argument of \[\left( \frac{-1-3i}{2+i} \right)={{\tan }^{-1}}\left( \frac{-1}{-1} \right)\] \[=225{}^\circ \] [ Since the given number lies on a 3rd quadrant ]
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