Question

If $z=1+\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}$ , then

• A. $amp\left(z\right)=\frac{-2\pi }{5}$
• B. $amp\left(z\right)=\frac{2\pi }{5}$
• C. $|z|=2\cos \frac{2\pi }{5}$
• D. $|z|=-2\cos \frac{2\pi }{5}$

Answer 1

Answer: (A), (C)

The correct options are
A $amp\left(z\right)=\frac{-2\pi }{5}$
C $|z|=2\cos \frac{2\pi }{5}$

$z=1+\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}$
$|z|=\sqrt{{(1+\cos \frac{6\pi }{5})}^2+{\left(\sin \frac{6\pi }{5}\right)}^2}$
$=\sqrt{2+2\cos \frac{6\pi }{5}}$
$=\sqrt{2(1+\cos \frac{6\pi }{5})}$
$=\sqrt{2\cdot 2{\cos }^2\left(\frac{3\pi }{5}\right)}$
$=2\left|\cos \left(\frac{3\pi }{5}\right)\right|$
$=-2\cos \left(\frac{3\pi }{5}\right)[\because \cos \frac{3\pi }{5}<0]$
$=2\cos \left(\frac{2\pi }{5}\right)$

$Re\left(z\right)=1+\cos \frac{6\pi }{5}>0$
$Im\left(z\right)=\sin \frac{6\pi }{5}<0$
So, $z$ lies in $4^{th}$ quadrant of Argand plane
$\tan \alpha =\frac{\left|\sin \frac{6\pi }{5}\right|}{|1+\cos \frac{6\pi }{5}|}$
$=\left|\frac{2\sin \frac{3\pi }{5}\cos \frac{3\pi }{5}}{2{\cos }^2\frac{3\pi }{5}}\right|$
$=\left|\tan \frac{3\pi }{5}\right|$
$=-\tan \frac{3\pi }{5}=\tan \frac{2\pi }{5}$
$\Rightarrow \alpha =\frac{2\pi }{5}$
Now, $amp\left(z\right)=\theta =-\alpha =-\frac{2\pi }{5}$

$\text{Alternate Solution}:$
$z=1+\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}$
$\Rightarrow z=1+\cos (\pi +\frac{\pi }{5})+i\sin (\pi +\frac{\pi }{5})$
$\Rightarrow z=1-\cos \frac{\pi }{5}-i\sin \frac{\pi }{5}$
$\Rightarrow z=2{\sin }^2\frac{\pi }{10}-i2\sin \frac{\pi }{10}\cos \frac{\pi }{10}$
$\Rightarrow z=2\sin \frac{\pi }{10}(\sin \frac{\pi }{10}-i\cos \frac{\pi }{10})$
$\Rightarrow z=2\cos (\frac{\pi }{2}-\frac{\pi }{10})[\cos (\frac{\pi }{2}-\frac{\pi }{10})-i\sin (\frac{\pi }{2}-\frac{\pi }{10})]$
$\Rightarrow z=2\cos \frac{2\pi }{5}[\cos \frac{2\pi }{5}-i\sin \frac{2\pi }{5}]$
$\Rightarrow z=2\cos \frac{2\pi }{5}[\cos (-\frac{2\pi }{5})+i\sin (-\frac{2\pi }{5})]$
$\therefore |z|=2\cos \frac{2\pi }{5}$ and $amp\left(z\right)=-\frac{2\pi }{5}$