in closed form. We prove below that in the case that the black hole mass exceeds the critical mass M$_{0}$ the roots of the equation g$_{00}$ ( r ) = 0 are simple. This result will play an important role in the following considerations. First we need to clarify the nature of the critical mass M$_{0}$ , which turns out to be the minimal mass to have horizons. Therefore, if we employ the horizon equation to define a function M = M ( r$_{H}$ ), where r$_{H}$ is a solution of the equation g$_{00}$ ( r ) = 0, we obtain Thus, having considered the derivative dM ( r$_{H}$ ) /dr$_{H}$ , we can look for r$_{0}$ such that the latter vanishes, i.e. dM/dr$_{H}$ | r 0 = 0. We do this in order to define M$_{0}$ ≡ M ( r$_{0}$ ). One can easily show that the above derivative vanishes if and only if From the properties of the gamma function one can easily derive the following result and as a consequence where r$_{m}$ ∈ [ 0 , r$_{H}$ ]. Eq. (5) can now be written as which admits a unique solution if and only if r$_{H}$ = r$_{m}$ = r$_{0}$ . We can conclude that there exists a unique horizon radius r$_{0}$ , corresponding to the critical mass M$_{0}$ . With this in mind we can establish the following lemma Lemma 1 Let M > M$_{0}$ and 0 < r − < r$_{+}$ such that g$_{00}$ ( r ± ) = 0 . Then, r − and r$_{+}$ are simple zeroes of (3). Proof. We give the proof for a generic solution r$_{H}$ of the horizon equation g$_{00}$ ( r$_{H}$ ) = 0. Therefore r$_{H}$ corresponds either to r$_{+}$ or r − . Notice that if the limit 4 lim r → r H g$_{00}$ ( r ) r − r$_{H}$ 1 4 θ 3 / 2 r$_{H}$ r 2 $_{m}$e − r 2 $_{m}$/ 4 θ = 1 4 θ 3 / 2 r 3 $_{H}$e − r 2 $_{H}$/ 4 $^{θ}$, (8) γ ( 3 / 2; r 2 $_{H}$/ 4 θ ) = 1 4 θ 3 / 2 r$_{H}$ r 2 $_{m}$e − r 2 $_{m}$/ 4 θ (7) γ ( 3 / 2; r 2 $_{H}$/ 4 θ ) = 1 4 θ 3 / 2 ∫ r H 0 dt t $^{2}$e − t $^{2}$/ 4 θ (6) γ ( 3 / 2; r 2 $_{H}$/ 4 θ ) = r$_{H}$ dγ dr$_{H}$ . (5) M ( r$_{H}$ ) ≡ √$_{π}$ 4 r$_{H}$ γ (3 / 2; r 2 $_{h}$/ 4 θ ) . (4)