is finite, then r$_{H}$ is a simple root. Since g$_{00}$ is differentiable on the interval [0 , ∞ ) we can expand it in a Taylor series and we obtain Hence, and in order to show that r$_{H}$ is a simple zero we need to prove that g ′ $_{00}$( r$_{H}$ ) = 0. Taking into account the fact that where a prime denotes differentiation with respect to the horizon radius and comparing the above equation with the Eq. (5), we can see that g ′ $_{00}$( r$_{H}$ ) vanishes if and only if r$_{H}$ = r$_{0}$ . This implies that M = M$_{0}$ which is at variance with the initial assumption. As a result we can conclude that g ′ $_{00}$( r$_{H}$ ) = 0 for M > M$_{0}$ and r$_{H}$ = r$_{0}$ . We show that the singularities of (2) can be removed by a suitable coordinate transfor- mation as in the case of the Reissner-Nordstr¨ om solution. In order to do that we shall follow [15]. Like in the Kruskal approach [14] we introduce coordinates u ( t, r ) and v ( t, r ) such that the original metric goes over to with the requirement that f 2 = 0. This will happen if u and v satisfy the non homogeneous system of first order nonlinear partial differential equations The next step is to find a suitable transformation of the variable r such that the above system becomes a homogeneous system of PDEs. If we multiply (11) by g 2 00 and we introduce a new 5 ∂$_{r}$u ∂$_{t}$u − ∂$_{r}$v ∂$_{t}$v = 0 . (12) f $^{2}$( u, v ) [ ( ∂$_{t}$v ) 2 − ( ∂$_{t}$u ) 2 ] = g$_{00}$ ( r ) , (10) 2 2 − 1 f $^{2}$( u, v ) ( ∂$_{r}$v ) − ( ∂$_{r}$u ) = − g 00 ( r ) , (11) ds 2 = f $^{2}$( u, v )( dv 2 − du $^{2}$) − r $^{2}$( u, v )( dϑ 2 + sin 2 ϑ dϕ $^{2}$) (9) g ′ $_{00}$( r$_{H}$ ) = 4 M √$_{πr}$$_{H}$ [ 1 r$_{H}$ γ ( 3 2 , r 2 H 4 θ ) − γ ′ ( 3 2 , r 2 H 4 θ )] lim r → r H g$_{00}$ ( r ) r − r$_{H}$ = g ′ $_{00}$( r$_{H}$ ) g$_{00}$ ( r ) = ( r − r$_{H}$ ) [ g ′ $_{00}$( r$_{H}$ ) + O ( r − r$_{H}$ ) ] . III. A NEW TRANSFORMATION