# Continuation to Euclidean BTZ Black hole

BTZ black hole in Lorentzian signature is given by $$ds^2= -fdt^2+f^{-1}dr^2+r^2(d\phi + N^{\phi} dt)^2$$ $$f=-M+r^2+\frac{J^2}{4r^2},~~~~~ N^{\phi}=-\frac{J}{2r^2}$$ $f$ can be wriitten as $$f=\frac{(r^2-r_+^2)(r^2-r_{-}^2)}{r^2}$$ where $$r_\pm = \sqrt{\frac{M}{2}} \bigg( 1\pm \sqrt{1-\frac{J^2}{M^2}}~ \bigg)^{\frac{1}{2}}$$

Analytic continuation to Euclidean signature is given by $t\rightarrow -i\tau,~ J\rightarrow iJ_E$:

$$ds^2= fd\tau^2+f^{-1}dr^2+r^2(d\phi + N_E^{\phi} d\tau)^2$$ $$f=-M+r^2-\frac{J_E^2}{4r^2},~~~~~ N^{\phi}=-\frac{J_E}{2r^2}$$

The $r_\pm$ becomes $$r_+ \rightarrow r^E_+= \sqrt{\frac{M}{2}} \bigg( 1+ \sqrt{1+\frac{J_E^2}{M^2}}~ \bigg)^{\frac{1}{2}}$$ $$r_- \rightarrow r_-^E= \sqrt{\frac{M}{2}} \bigg( 1- \sqrt{1+\frac{J_E^2}{M^2}}~ \bigg)^{\frac{1}{2}}$$

I found that in most papers $r_-$ is set to be $r_-^E= -i \vert r_-^E\vert$.

My question is why not $r_-^E= i \vert r_-^E\vert$. In this case the relation $J=2r_+r_-$ beomes $J^E=2r^E_+ \vert r^E_- \vert$ under continuation.