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

$\text{ Let }z=(1+\cos 2\alpha )+i\sin 2\alpha$

$=2{\cos }^2\alpha +2i\sin \alpha \cos \alpha$

$=2\cos \alpha (\cos \alpha +i\sin \alpha )$

$=2\cos \alpha (\cos \alpha +i\sin \alpha )$

$=-2\cos \alpha [\cos (\alpha -\pi )+i\sin (a-\pi \left)\right]\left(\because \frac{\pi }{2}<\alpha <\frac{3\pi }{2}\right)$

$\therefore \left|z\right|=-2\cos \alpha \text{ and }\mathrm{Arg}\left(z\right)=\alpha -\pi$