Why does $|\exp{(R^2 i(\cos{2t} +i \sin{2t})}| = \exp{ (-R^2 \sin{2t})} $?

Question

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?

Answer 1

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

Answer 2

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 $\theta=R^2cos(2t)+2k\pi$.