Let $x = -\frac{(2\;W({\pi\over2}))}{\pi}$, where $W$ denotes the Lambert W-function.
As $${\log(i^2)\over i} = \pi$$ and $${\log(x^2)\over x}=\pi$$
Does $x = i$?
2026-03-27 19:31:38.1774639898
Does $i = -\frac{(2\;W({\pi\over2}))}{\pi}$
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Hint:assume $x=a+ib$ then $${\log(x^2)\over x}=\pi\to \pi(a+ib)=\log(x^2)=\ln(|x^2|)+i\arg(x^2)=2\mathrm{Ln}(|x|)+i2\arg(x)$$$$\to \begin{cases} 2\ \ln(|\sqrt{a^2+b^2}|)=\pi a \\ \pi b=2\arg(x) \\ \end{cases}$$