What is the real part of $ e^{e^{i \theta}} ?$

Question

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.

Answer 1

$$z = e^{e^{i \theta} }$$

$$\implies z=e^{\cos \theta +i\sin \theta }$$

$$\implies z=e^{\cos \theta} e^{i\sin \theta }$$

$$\implies z=e^{\cos \theta}(\cos (\sin\theta) +i\sin(\sin \theta))$$

Answer 2

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.

Answer 3

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 $$z=e^{e^{i\theta}}=e^{cos\theta +isin\theta}=e^{cos\theta}[\cos (\sin\theta) +i\sin(\sin \theta)]$$