part of this equation reads $$\begin{aligned}
\operatorname{Re}{(\omega_n)} \Bigl(2 \operatorname{Im}{(\omega_n)}A^2+ B^2\Bigr)=0\,.\end{aligned}$$ Therefore, if the black hole is unstable (i.e., $\operatorname{Im}{(\omega_n)}>0$) then $\operatorname{Re}{(\omega_n)}=0$. The unstable modes do not oscillate. So, the unstable modes are equivalent to the condition $\omega_n^2<0$ (i.e., purely imaginary modes), which from (\[eq:master2\]) concludes that the stability of the spacetime is related to the positivity of the operator $-\frac{d^2}{dr^{*2}}+V$.

We now turn to the real part of (\[eq:AB\]) which reads $$\begin{aligned}
\label{eq:AB2}
\int_{\mathbb{R}}{\rm d}r^* \Bigl[\Bigl |\frac{d\psi_n}{dr^*}\Bigr |^2+V |\psi_n |^2\Bigr]&=(\operatorname{Re}{(\omega_n)}^2-\operatorname{Im}{(\omega_n)}^2)A^2\nonumber\\
&-\operatorname{Im}{(\omega_n)}B^2\,.\end{aligned}$$ If the mode is unstable, we have $\operatorname{Re}{(\omega_n)}=0$ and $\operatorname{Im}{(\omega_n)}>0$, which implies the right-hand side of eq. (\[eq:AB2\]) being negative, and therefore this condition can be fulfilled only if the potential on the left-hand side of eq. (\[eq:AB2\]) is negative. Of course, a negative potential can yet give a positive integral. On the contrary, if the potential is positive, we necessarily have stability. In the case of a negative potential, the “S-deformation” approach is useful. For that, we introduce a smooth function and a new “derivative”, $D=\frac{d}{dr^*}+S$. We get, after integration by parts $$\begin{aligned}
\label{eq:fin}
\int_{\mathbb{R}}{\rm d}r^* \Bigl[\Bigl |\frac{d\psi_n}{dr^*}\Bigr |^2+V |\psi_n |^2\Bigr]&=\int_{\mathbb{R}}{\rm d}r^* \Bigl[ |D\psi_n |^2+W |\psi_n |^2\Bigr]\nonumber\\
&-S|\psi_n |^2\Bigl |_{r^{*}=-\infty}^{r^{*}=+\infty}\,,\end{aligned}$$ where the new potential $W$ is defined as 