I want to prove that $ \left|e^{2iRe^{it}}\right| = e ^{ -2R\sin(t)} $ for some constant R.
I know that $ e ^{ -2R\sin(t)}= e^{-2R\times\text{Im}(e^{it})} $ where Im denotes the imaginary part, so i just need to prove that
$$ \left|e^{2iRe^{it}}\right| = e^{-2R\times\text{Im}(e^{it})} $$
but i can not find the relation between the complex modulus and the imaginary part.
Best Answer
Hint: $\left| e^{x+iy} \right| = \left| e^{x} \right| \left| \cos y + i\sin y\right| = e^x$ for $x, y \in \mathbb{R}$.