Let $W=h(z)$
$h(z) = z\uparrow\uparrow n$ when n approaches to infinity.
its clear that:
$z^{h(z)} = h(z)$ then $z^w = w$
$w = \frac1 {(ssrt(1/z))}$ where $ssrt(x)$ is super square root of $x$.
$h(z) = \frac1 {(ssrt(1/z))}$
The super square root $ssrt(x)$ has one real value when $x > 1$, two real values for $e^{-\frac{1}{e}} < x < 1$ and imaginary values when $x < e^{-\frac{1}{e}}$.
Example:
$z=\sqrt{2}$
$h(\sqrt{2}) = \frac1 {ssrt\left(\frac1{\sqrt{2}}\right)}$
Note that $(\frac12)^{\frac12}=\frac1{\sqrt{2}}$ and also $(\frac14)^{\frac14}=\frac1{\sqrt{2}}$
Then $h(z)=h(\sqrt{2})=\frac1{\frac12}=2$
and also $h(z)=h(\sqrt{2})=\frac1{\frac14}=4$
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