What is the real part of $ e^{e^{i \theta}} ?$ How to find the real part of the complex number (in Euler's form) $ z = e^{e^{i \theta } } $ ?
I got confused on how to proceed.
I am a beginner to complex numbers.
 A: $$z = e^{e^{i \theta} }$$
$$\implies z=e^{\cos \theta +i\sin \theta }$$
$$\implies z=e^{\cos \theta} e^{i\sin \theta }$$
$$\implies z=e^{\cos \theta}(\cos (\sin\theta) +i\sin(\sin \theta))$$
A: We have, $$e^{e^{i\theta}} = e^{\cos \theta + i\sin \theta} = e^{\cos \theta} \times e^{i\sin \theta} = e^{\cos \theta} [\cos(\sin \theta) + i \sin(\sin \theta)]$$
This comes due to using Euler’s formula twice (In expanding $e^{i\theta}$ and in expanding $e^{i\sin \theta}$) . 
Hope you can take it from here.
A: For Euler's formula:
$z=x+iy=r(cos\theta+isin\theta)$ (It is easy part.)
$cos\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+...$
$sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$| Then 
$cos\theta+isin\theta=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!}+...+i(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...)$$\hspace{56pt}= 1+i\theta+(\frac{(i\theta)^2}{2!}+(\frac{(i\theta)^3}{3!}+(\frac{(i\theta)^4}{4!}...$ $\hspace{56pt}= e^{i\theta}$. (Because $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$)Hence $$z=e^{e^{i\theta}}=e^{cos\theta +isin\theta}=e^{cos\theta}[\cos (\sin\theta) +i\sin(\sin \theta)]$$
