Question

The principal value of the $arg\left(z\right)$ and $|z|$ of the complex number

$z=1+\cos \left(\frac{11\pi }{9}\right)+i\sin \left(\frac{11\pi }{9}\right)$ are respectively

• A. $\frac{11\pi }{18},2\cos \frac{\pi }{18}$
• B. $-\frac{7\pi }{18},2\cos \frac{7\pi }{18}$
• C. $\frac{2\pi }{9},2\cos \frac{7\pi }{18}$
• D. $-\frac{\pi }{9},-2\cos \frac{\pi }{18}$

Answer 1

Answer: (B)

Given:

$z=1+\cos \left(\frac{11\pi }{9}\right)+\sin \left(\frac{11\pi }{9}\right)$

$=2{\cos }^2\left(\frac{11\pi }{18}\right)+i\times 2\sin \left(\frac{11\pi }{18}\right)\cos \left(\frac{11\pi }{18}\right)$

$=2\cos \left(\frac{11\pi }{18}\right)[\cos \frac{11\pi }{18}+i\sin \frac{11\pi }{18}]$

Now it is in the form of $\left|z\right|[\cos \theta +i\sin \theta ]$

$\Rightarrow \left|z\right|=2\cos \left(\frac{11\pi }{18}\right)$

$\Rightarrow {90}^{\circ }<\frac{11\pi }{18}<{180}^{\circ }$

$\therefore$ Principle argument $=\pi -\frac{11\pi }{18}=\frac{7\pi }{18}$

Hence, the answer is $\frac{7\pi }{18},2\cos \left(\frac{11\pi }{18}\right).$

Hence, option B is correct.