QED Branch cut

função que deve ficar negativa

f[x_,p2_]=(1-x)m0^2+x μ^2-x(1-x)p2
Collect[Expand[%],x]

(* m0->10,μ->1  -> p2branch = (1+11)^2  *)
Plot3D[f[x,p2]/.{m0->10,μ->1},{x,0,1},{p2,-200,300},PlotRange->{Automatic,{-100,300},{Automatic,0}},AxesLabel-> {x,p2,f}]

Animate[Plot[f[x,p2]/.{m0->10,μ->1},{x,0,2},AxesLabel-> {x,f},ImageSize->600,PlotRange->{{0,2},{-40,100}},PlotStyle->Thick],{p2,40,300},AnimationRunning->False]

(* m0->10,μ->1  -> p2branch = (1+11)^2  *)
Plot3D[f[x,p2]/.{m0->10,μ->1},{x,0,1.2},{p2,-200,200},PlotRange->{Automatic,{-100,150},{-10,0}}];

posição do mínimo

fp[x_,p2_]=D[f[x,p2],x]
x0[p2_]=x/.Solve[fp[x,p2]==0,x][[1]]

valores de p nos quais os mínimos estao abaixo de zero

f[x0[p2],p2];
Collect[FullSimplify[%],p2]
Solve[%==0,p2]
p2p=p2/.%[[2]];
p2m=p2/.%%[[1]];

posiçao do mínimos em x

Simplify[x0[p2m]]

Simplify[x0[p2p]]