A) \[60{}^\circ \]
B) \[120{}^\circ \]
C) \[240{}^\circ \]
D) \[300{}^\circ \]
Correct Answer: C
Solution :
\[z=\left( \frac{1-i\sqrt{3}}{1+i\sqrt{3}} \right)\] \[\Rightarrow \] \[z=\frac{1-i\sqrt{3}}{1+i\sqrt{3}}\times \frac{1-i\sqrt{3}}{1-i\sqrt{3}}\] \[\Rightarrow \] \[z=\frac{1-3-2\sqrt{3}i}{1+3}=\frac{-2-2\sqrt{3}i}{4}\] \[\Rightarrow \] \[z=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\] \[\Rightarrow \] \[z=\cos 240{}^\circ -i\,\sin 240{}^\circ \] Thus, \[\arg \,(z)=240{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec
