Question

If $\frac{\pi }{2}<\alpha <\frac{3\pi }{2}$ then the modulus argument form $(1+cos2\alpha )+isin2\alpha$ is :

• A. $2cos\alpha \left[cos\right(\pi +\alpha )+isin(\pi +\alpha \left)\right]$
• B. $2cos\alpha [cos\alpha +isin\alpha ]$
• C. $2cos\alpha \left[cos\right(-\alpha )+isin(-\alpha \left)\right]$
• D. $-2cos\alpha \left[cos\right(\pi -\alpha )+isin(\pi -\alpha \left)\right]$

Answer 1

The correct option is A $2cos\alpha \left[cos\right(\pi +\alpha )+isin(\pi +\alpha \left)\right]$

$z=1+\cos 2\alpha +i\sin 2\alpha$
$\left|z\right|=\sqrt{{(1+\cos 2\alpha )}^2+\left({\sin }^{2}2\alpha \right)}$
$=\sqrt{1+{\cos }^{2}2\alpha +2\cos 2\alpha +{\sin }^{2}2\alpha }$
$=\sqrt{1+1+2\cos 2\alpha }$
$=\sqrt{2(1+\cos 2\alpha )}$
$=\sqrt{2(1+2{\cos }^2\alpha -1)}$
$=2\cos \alpha$
$argz=\varnothing ={\tan }^{-1}\left(\frac{\sin 2\alpha }{1+\cos 2\alpha }\right)={\tan }^{-1}\left(\frac{\sin 2\alpha }{2{\cos }^2\alpha }\right)$
$={\tan }^{-1}\frac{2\sin \alpha \cos \alpha }{2{\cos }^2\alpha }$
$={\tan }^{-1}\tan \alpha$
$=(\alpha +\pi )$
$\to$ As $\alpha$ lies in 2nd and 3rd quadrant,

$\therefore$ The modulus argument form of the given complex number is:

$2\cos \alpha \left[\cos \right(\pi +\alpha )+i\sin (\pi +\alpha \left)\right]$