Is it possible to break $f(z)=e^{z^e}$ into real and complex parts? Example: $h(z)=z^2$
$$h(z)=f(x+iy)=(x+iy)^2$$
$$=x^2-y^2+i(2xy)$$
I am wondering if I could do the same for $f(z)=e^{z^e}$. Any help would be greatly appreciated. Thanks.
 A: The problem with $z^e$ is that since $e$ is irrational, this function has infinitely many values. If $z=r(\cos\varphi + i\sin\varphi)$,
$$
z^e = r^e(\cos(e\varphi+2e\pi n)+i\sin(e\varphi+2e\pi n)), \qquad n\in\mathbb Z
$$
If you are ok with that, then
$$
e^{z^e} = e^{r^e\cos(e\varphi+2e\pi n)}\Big(\cos\left(r^e\sin(e\varphi+2e\pi n)\right)
+i\sin\left(r^e\sin(e\varphi+2e\pi n)\right)\Big),\qquad n\in\mathbb Z
$$
If we want to choose principal branch of $\log z$, then $\varphi=\arg z$ and $n=0$.
A: First, remember that $w^z = e ^ {z \log w}$, for complex numbers, is not defined unambiguously because that depends on the choice of the logarithm. Let us agree to use the principal value as the logarithm. Then we have:
$$z^e = e^{e \log z} = e^{e (\log r + i \theta)}$$
where, for the sake of simplicity, I used polar coordinates.
Remembering Euler's fomula, last expression is equal to $$e^{e \log r}(\cos{e\theta} + i \sin{ e\theta})$$
Now, $$e^{z^e}=e ^ {e^{e \log r} (\cos{e \theta} + i \sin {e \theta})}$$
and, using again Euler's formula, we get:
$$e^{e^{e \log r} \cos{e \theta}}\bigg(\cos{(e^{e \log r}\sin{e\theta}) + i \sin(e^{e \log r} \sin {e\theta})\bigg)}$$
A: Maple says
$$
\exp\left(z^\rm{e}\right)
={{\rm e}^{ \left( {x}^{2}+{y}^{2} \right) ^{{\rm e}/2}\cos \left( {
\rm e}\arctan \left( y,x \right)  \right) }}\cos \left(  \left( {x}^{2
}+{y}^{2} \right) ^{{\rm e}/2}\sin \left( {\rm e}\arctan \left( y,x
 \right)  \right)  \right) 
\\+i{{\rm e}^{ \left( {x}^{2}+{y}^{2}
 \right) ^{{\rm e}/2}\cos \left( {\rm e}\arctan \left( y,x \right) 
 \right) }}\sin \left(  \left( {x}^{2}+{y}^{2} \right) ^{{\rm e}/2}
\sin \left( {\rm e}\arctan \left( y,x \right)  \right)  \right) 
$$
