Imaginary exponents and bases

When $n=-1$ , then

$\begin{aligned} z^{-1} & = \log_z (-1) = \frac {\ln (-1)}{\ln z} = \frac {\ln (e^{i\pi}}{\ln z} = \frac {i\pi}{\ln z} \end{aligned}$

$\begin{aligned} \implies z^{-1} \ln z & = i \pi & \small \blue{\text{Let }x = \ln z \implies e^x = z} \\ e^{-x} \cdot x & = i \pi \\ -x e^{-x} & = - i \pi \\ W (-xe^{-x}) & = W(-i\pi) & \small \blue{\text{See reference}} \\ \implies -x & = W (-i\pi) \\ - \ln z & = W(-i\pi) & \small \blue{\text{Note that }z^{-1}\ln z = i\pi} \\ -i\pi z & = W(-i\pi) \\ \implies z & = \frac {iW(-i\pi)}\pi \end{aligned}$

Therefore $a = \pi$ and $|\sqrt {\lfloor a \rfloor} | \approx \boxed{1.73}$ .

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Reference: What is Lambert W-function?

$\ln (-1)=iπ\implies a=π\implies |\sqrt {\lfloor a\rfloor}|=\sqrt 3\approx \boxed {1.732050808}$ .