Remove the blank of branch cuts

Question

H = {{I*(x + I*y), 1}, {1, -I*(x + I*y)}};
{x1, x2} = Eigenvalues[H];

Plot3D[{Im[x1], Im[x2]}, {x, 0, 2}, {y, -1, 1}, PlotStyle -> {LightRed, LightRed}, Mesh -> 0]

There is a white blank on the surface because of the branch cut. How to remove this "white line" to make the two surfaces as a whole part?

Answer 1

The desired imaginary parts are represented by v in algebraic formulation below. Put u before v in the variables for GroebnerBasis to get the first polynomial to define v in terms of x and y. Reverse (v before u, both before x, y) to get the real part u in terms of x and y.

branches = u + I v - Eigenvalues[H];
surf = Times @@ branches // Simplify;
parts = ReIm[surf] // ComplexExpand;
gb = GroebnerBasis[parts, {u, v, x, y}];
ContourPlot3D[
 First[gb] == 0 // Evaluate, {x, 0, 2}, {y, -1, 1}, {v, -2, 2},
 Mesh -> None, ContourStyle -> LightRed, MaxRecursion -> 6, 
 PlotPoints -> 20]

Added alternative:

Another way that happens to remove the branch-cut jump (the use of polar coordinates is similar to Henrik's, but it's the combination with ComplexExpand that moves the branch cut to the boundary of the plot):

imEVs = FullSimplify[
   ComplexExpand[Im@Eigenvalues[H], TargetFunctions -> {Re, Im}],
   x > 0 && -1 < y < 1];
ParametricPlot3D[
 Evaluate@Thread[{x, y, imEVs}],
 {x, 0, 2}, {y, -1, 1}, PlotStyle -> LightRed, Mesh -> None, 
 Exclusions -> None, BoxRatios -> {1, 1, 1}
 ]

Answer 2

Using polar coordinates helps a little

plot = With[{
   opts = 
    Sequence[PlotStyle -> {LightRed, LightRed}, Mesh -> 0, 
     PlotPoints -> {73, 20},
     BoundaryStyle -> None,
     PlotRange -> {{0, 2}, {-1, 1}, {-2, 2}}
     ],
   ϵ = 1. 10^-6,
   aa = {x, y, Im[x1]} /. {x -> 1 + r Cos[t], y -> r Sin[t]},
   bb = {x, y, Im[x2]} /. {x -> 1 + r Cos[t], y -> r Sin[t]}
   },
  Show[
   ParametricPlot3D[aa, {t, ϵ, 2 Pi - ϵ}, {r, 0, 2},
     opts],
   ParametricPlot3D[bb, {t, ϵ, 2 Pi - ϵ}, {r, 0, 2},
     opts]
   ]
  ]

However, the slit can still be seen from different view points. But one can try and glue the two GraphicsComplexes within plot together in order to fix this.

Answer 3

Somewhat inelegant, but works?

Plot3D[{Max[Im[x1], Im[x2]], Min[Im[x1], Im[x2]]}, {x, 0, 2}, {y, -1, 1},
  PlotStyle -> {LightRed, LightRed}, Mesh -> 0, ExclusionsStyle -> LightRed]

Answer 4

Just for fun note the eigenvalue expression is: $\lambda^2=1-(x^2+2ixy-y^2)=1-(x+iy)^2$. Letting $z=x+iy$ we have the expression $\lambda=\sqrt{1-z^2}$

So basically, $\lambda$ is just the square root function offset to x=1. And one easy way to plot this is:

myf[z_] := Sqrt[1 - z^2];
p1 = ParametricPlot3D[{Re[z], Im[z], Im@myf[z]} /. 
   z -> 1 + r Exp[I t], {r, 0, 1}, {t, 0, 2 \[Pi]}]
p2 = ParametricPlot3D[{Re[z], Im[z], -Im@myf[z]} /. 
   z -> 1 + r Exp[I t], {r, 0, 1}, {t, 0, 2 \[Pi]}]
Show[{p1, p2}]