$|\exp(iRe^{i\theta})|=\exp(-R\sin\theta)$?

Question

I'm reading Conway's complex analysis book and on page 116 he writes:

Why is the last equality true?

Answer 1

$ |e^{iRe^{i\theta}}| = |e^{{iR(\cos\theta + i\sin\theta)}}| = |e^{{iR\cos\theta}}e^{-R\sin\theta}| = |e^{-R\sin\theta}|, $

since $R\cos\theta$ is real, hence $|e^{{iR\cos\theta}}| =1$

Answer 2

Because for $\;x,y\in\Bbb R\;$ :

$$e^{x+iy}=e^xe^{iy}\implies\left|e^{x+iy}\right|=|e^x|\,|e^{iy}|=e^x$$

since $\;|e^{iy}|=|\cos x+i\sin x|=\sqrt{\cos^2x+\sin^2x}=1\;$

And in your case:

$$e^{iRe^{i\theta}}=e^{iR(\cos\theta+i\sin\theta)}=e^{-R\sin\theta}e^{iR\cos\theta}$$