If $\frac{\pi }{2}<\theta <\frac{3\pi }{2}$, the modulus and argument form of $(1+\cos 2\theta )+i\sin 2\theta$ is 

If $z=r{e}^{i\theta }\text{ then }r=+\text{ ive and }\theta \in [-\pi ,\pi ]$

$E=2\cos \theta (\cos \theta +i\sin \theta )$

Since $\frac{\pi }{2}<\theta <\frac{3\pi }{2}\therefore r=2\cos \theta =-\text{ ive }$

$\begin{array}{r}\therefore E=(-2\cos \theta )(-\cos \theta -i\sin \theta )\\ E=-2\cos \theta [\cos (\pi +\theta )+i\sin (\pi +\theta \left)\right]\end{array}$

where $\pi +\frac{\pi }{2}<\pi +\theta <\pi +\frac{3\pi }{2}$

$\therefore \frac{3\pi }{2}<\pi +\theta <\frac{5\pi }{2}$

$\text{ i.e.arg }(\pi +\theta )\mathrm{lies}\mathrm{in}\left(\frac{3\pi }{2},\frac{5\pi }{2}\right)$

$\text{ or in }\left(\frac{3\pi }{2}-2\pi ,\frac{5\pi }{2}-2\pi \right)\text{ i.e., in }\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

which is a subset of $(-\pi ,\pi )$.