Why does $|\exp{(R^2 i(\cos{2t} +i \sin{2t})}| = \exp{ (-R^2 \sin{2t})} $? why does |$\exp{(R^2 i(\cos{2t} +i \sin{2t})}$| $= \exp{ (-R^2 \sin{2t})} $
From the question I'm thinking that $i \cos{2t} =0$ but I'm not sure why?
 A: This equality can be proven in a less complicated way. Let
$Z:=\exp(R^2 i(\cos{2t} +i \sin{2t}))=\exp{(R^2 i\cos{2t} -R^2 \sin{2t})}$
$=\exp(-R^2 \sin{2t}) \times \exp(i R^2\cos{2t})$ where we recognize the form $\rho e^{i\theta}$ (polar form of complex number $Z$). By unicity of the polar form, we have
$$\rho=|Z|=\exp(-R^2 \sin{2t})$$
as desired, and $\theta=R^2cos(2t)+2k\pi$.
A: Notice:


*

*$$i\left(\cos(2t)+\sin(2t)i\right)=\cos(2t)i+\sin(2t)i^2=\cos(2t)i-\sin(2t)$$


So, we get the following identity:


*

*$$\exp\left[i\text{R}^2\left(\cos(2t)+\sin(2t)i\right)\right]=\exp\left[\text{R}^2\cos(2t)i-\text{R}^2\sin(2t)\right]=$$
$$\exp\left[\text{R}^2\cos(2t)i\right]\exp\left[-\text{R}^2\sin(2t)\right]$$


Now, we get that:
$$\left|\exp\left[\text{R}^2i\left(\cos(2t)+\sin(2t)i\right)\right]\right|=\left|\exp\left[\text{R}^2\cos(2t)i-\text{R}^2\sin(2t)\right]\right|=$$
$$\left|\exp\left[\text{R}^2\cos(2t)i\right]\exp\left[-\text{R}^2\sin(2t)\right]\right|=\left|\exp\left[\text{R}^2\cos(2t)i\right]\right|\left|\exp\left[-\text{R}^2\sin(2t)\right]\right|=$$
$$\exp\left[\Re\left[\text{R}^2\cos(2t)i\right]\right]\exp\left[\Re\left[-\text{R}^2\sin(2t)\right]\right]=\exp\left[0\right]\exp\left[-\text{R}^2\sin(2t)\right]=$$
$$1\exp\left[-\text{R}^2\sin(2t)\right]=\exp\left[-\text{R}^2\sin(2t)\right]$$
So, your equality is right, when we are talking about the absolute value
