Question
The principle value of the $$\arg \left( z \right)$$ and $$\left| z \right|$$ of the complex number $$z = 1 + \cos \left( {\frac{{11\pi }}{9}} \right) + i\sin \left( {\frac{{11\pi }}{9}} \right)$$ are respectively.
A.
$$\frac{{11\pi }}{8},2\cos \left( {\frac{\pi }{{18}}} \right)$$
B.
$$ - \frac{{7\pi }}{18}, - 2\cos \left( {\frac{11\pi }{{18}}} \right)$$
C.
$$\frac{{2\pi }}{9},2\cos \left( {\frac{7\pi }{{18}}} \right)$$
D.
$$ - \frac{{\pi }}{9}, - 2\cos \left( {\frac{\pi }{{18}}} \right)$$
Answer :
$$ - \frac{{7\pi }}{18}, - 2\cos \left( {\frac{11\pi }{{18}}} \right)$$
Solution :
$$z = 1 + \cos \frac{{11\pi }}{9} + i\sin \frac{{11\pi }}{9}$$
$$\operatorname{Re} \left( z \right) > 0$$ and $$\operatorname{Im} \left( z \right) < 0,$$ so the number lies in the fourth quadrant. Also
$$\eqalign{
& z = 2\cos \frac{{11\pi }}{{18}}\left\{ {\cos \frac{{11\pi }}{{18}} + i\sin \frac{{11\pi }}{{18}}} \right\} \cr
& = 2\cos \frac{{11\pi }}{{18}}\left\{ {\cos \left( { - \frac{{7\pi }}{{18}}} \right) + i\sin \left( { - \frac{{7\pi }}{{18}}} \right)} \right\} \cr
& \therefore \arg \left( z \right) = - \frac{{7\pi }}{{18}} \cr
& \left| z \right| = \left| {2\cos \frac{{11\pi }}{{18}}} \right| = - 2\cos \frac{{11\pi }}{{18}} \cr} $$