This might be a start. Define the expression, then use ComplexExpand to form explicit real and imaginary parts.

g = 1/10;
expr = 
  Tanh[Sqrt[-x^2 - y^2]*3/10] (y^2 + I*g*y - 4) Sqrt[-x^2 - 
      y^2] + (y^2 + I*g*y) Sqrt[-x^2 - 
      y^2 ((y^2 + I*g*y - 4)/(y^2 + I*g*y))];

e2 = expr /. x -> s + I*t;
{re, im} = ComplexExpand[{Re[e2], Im[e2]}];

 ContourPlot[(re /. y -> 0.7) == 0, {s, -10, 10}, {t, -10, 10}]

ContourPlot[(im /. y -> 0.7) == 0, {s, -10, 10}, {t, -10, 10}]

Use ContourPlot3D, it should give you your result; a 3d surface.

You should do some debugging. Look hard at the messages, the NDSolveValue result, that sort of thing. Maybe even start in a fresh kernel.

Ok, thanks!

Dear Daniel, thank you for you help. Could you tell me what version of Mathematica do you use (because I can't use this code in my version 7)?

No, no, no. Dear Daniel, I am sorry that my explanations were probably unclear. I need exactly what I have shown you in my version (because I know the answer for this particular example). Now I attach a picture where I explain everything in symbols (and in a couple of words). I think that this picture makes everything much more clear without words (I am sorry that I have presented in just on paper, but I think it is clear enough).

However, it would good to try to explain again what I need by words. I have a function f[x,y], where y is real and x is complex. I need to solve this equation for y from 0 to 10, i.e. to find x for every y from 0 to 10. After getting the solutions (Re(x), Im(x), y)I need to plot two graphs: y as a function of Re(x) and y as a function of Im(x) (i.e. use the numbers from 1st and 3d column for the first graph and the numbers from 2nd and 3d column for the second graph)

Attachments:

Dear Daniel. I have decided to check your method I have considered a very simple problem (x=f[y]) which can be easily solved by a very straightforward method (just using Plot). I don't know why but the results are essentially different.

I attach my file, but the general idea is very simple. If we have x = f[y] and x=x'+ix'', we can just plot x'=Re[f[y]] and x''=Im[f[y]] and compare these graphs with the graphs which we get using your method.

Attachments:

Your nb gave a nice plot when I executed the top cell. Maybe you had other definitions active at the time that interfered in some way?

Thank you. But can I further extract two 2D graphs from this 3D graph? I mean I can plot F[Re[x],Im[x],y]=0, but I would like to get two projections of this graph: Re[x],y and Im[x],y.