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

$$\begin{gathered} 1+\cos 2\alpha +i \sin 2\alpha \\ =2{\cos }^2\alpha +2i \sin \alpha \cos \alpha \\ =2\cos \alpha (\cos \alpha +i\mathrm{si}n \alpha ) \\ =-2\cos \alpha \left(\right(\cos (\pi +\alpha )+i \sin (\pi +\alpha )) \\ observe that \cos \alpha <0 \end{gathered}$$