Complex exercise about nested power i's solved by two method one on them is Lambert function

Exercise:

Let $z\in \mathbb{C}$ such that $z= { i }^{ { i }^{ i^{.^{.^{.}}} } } $Find the value of ${{\left| z \right|}^{2}}$

Solution: We have $z= { i }^{ { i }^{ i^{.^{.^{.}}} } }$hence $z={{i}^{z}}={{\left( i \right)}^{z}}$

So $z={{e}^{\ln \left( {{i}^{z}} \right)}}={{e}^{z\ln i}}$ but $\ln \left( i \right)=\ln 1+i\left( \frac{\pi }{2} \right)=i\left( \frac{\pi }{2} \right)$

Hence $z={{e}^{z\frac{\pi }{2}i}}={{e}^{\frac{z\pi }{2}i}}\Rightarrow 1=\frac{{{e}^{\frac{z\pi }{2}i}}}{z}\Rightarrow 1=\frac{1}{z{{e}^{\frac{-z\pi }{2}i}}}\Leftrightarrow z{{e}^{\frac{-\pi }{2}zi}}=1$

Put $w=-\frac{\pi }{2}zi\Leftrightarrow 2w=-\pi zi\Rightarrow z=-\frac{2w}{\pi i}$

Hence $z{{e}^{\frac{-\pi }{2}zi}}=1\Leftrightarrow -\frac{2w}{\pi i}{{e}^{w}}=1\Leftrightarrow -2w{{e}^{w}}=\pi i\Leftrightarrow w{{e}^{w}}=-\frac{\pi }{2}i$

So $W\left( z \right){{e}^{W\left( z \right)}}=z$ where $z=-\frac{\pi }{2}i$ and $W$ is the Lambert Function

Thus $w=W\left( -\frac{\pi }{2}i \right)\Rightarrow z=-\frac{2w}{\pi i}=\frac{{{i}^{2}}2w}{\pi i}=\frac{2wi}{\pi }=\frac{2i}{\pi }W\left( -\frac{\pi }{2}i \right)=0.438283+0.360592i$

Therefore ${{\left| z \right|}^{2}}={{\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}={{a}^{2}}+{{b}^{2}}={{\left( 0.438283 \right)}^{2}}+{{\left( 0.360592 \right)}^{2}}=0.3221188633963876$

Remark that from lambert function that $ { z }^{ { z }^{ z^{.^{.^{.}}} } }=\frac{W\left( -\ln z \right)}{-\ln z}$

So $ { i }^{ { i }^{ i^{.^{.^{.}}} } }=\frac{W\left( -\ln i \right)}{-\ln i}=0.4382829367270321+0.3605924718713855i$

We can solve this exercise also by using Elementary properties for complex numbers and some of numerical analysis  by supposing $z=a+ib\Rightarrow z={{i}^{z}}={{i}^{a+ib}}$ but $i={{e}^{i\frac{\pi }{2}}}$

So $z={{\left( {{e}^{i\frac{\pi }{2}}} \right)}^{a+ib}}={{e}^{i\frac{a\pi }{2}}}{{e}^{\frac{-b\pi }{2}}}=r{{e}^{i\theta }}$ thus $r={{e}^{-\frac{b\pi }{2}}}\,\,\And \,\,\theta =\frac{a\pi }{2}$ where $\theta =\arctan \left( \frac{b}{a} \right)$

So ${{r}^{2}}={{e}^{-b\pi }}\,\,\And \,\arctan \left( \frac{b}{a} \right)=\frac{a\pi }{2}\Leftrightarrow \frac{b}{a}=\tan \left( \frac{a\pi }{2} \right)\Leftrightarrow b=a\tan \frac{a\pi }{2}$

But ${{r}^{2}}={{a}^{2}}+{{b}^{2}}={{e}^{-b\pi }}\Rightarrow {{a}^{2}}+{{a}^{2}}{{\tan }^{2}}\left( \frac{a\pi }{2} \right)={{e}^{-a\tan \left( \frac{a\pi }{2} \right)\pi }}$

Now just solving to find $a\,\And \,b$ using numerical method