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.

Dear Daniel, thank you so much for your help! Hovewer, I am not sure that this is exactly what I need, because I don't quietly understand how it works.

I can explain what I need in other words. I need the graph of the equation F[x,y]=0, x is a complex number, y is a real number, {y,0,8}. The graph of this equation is 3D, because we have 3 variables Re[x], Im[x], y. I need two 2D graphs y as a function of Re[x] and Im[x], I mean that I need two projection of this 3D graph.

Dear Daniel, could you clarify how do you do this? I try to use your code, but I don't get a result.g = 1/10; expr = x - Sqrt[y^2 + I*y]; e2 = expr /. x -> s + I*t; {re, im} = ComplexExpand[{Re[e2], Im[e2]}]; root[yval_?NumericQ] := {s, t} /. FindRoot[{re == 0, im == 0} /. y -> yval, {s, RandomReal[{-5, 5}]}, {t, RandomReal[{-5, 5}]}] pts = Quiet[Table[root[j], {j, .01, 8, .01}]]; ListPlot[pts]

Attachments:

Dear Daniel, I am sorry for interruption. I don't know what was the problem, but now everything is fine. Could you answer a couple of other questions? Firstly, what shoul I do to consider this problem only in the region {Im[x],0,8}, {Re[x],0,8}, {y,0,8}. And how can I distinguish between the dependense of y on Re[x] and y on Im[x] depicted on the figure? Also, could you clarify what do I change when alter RandomReal[{0, 5}] and {j, .01, 5, .01}?

Attachments:

The plots look more similar if you recognize that the equivalent form for the second one should use ParametricPlot instead of Plot (such distinctions are very important). The ListPlot will need a range setting as well. Try the two below.

ListPlot[pts, PlotRange -> All]

ParametricPlot[{Re[y*Sqrt[(y^2 + 1/10*I*y - 4)/(2 y^2 + 1/10*2*I*y - 4)]], 
  Im[y*Sqrt[(y^2 + 1/10*I*y - 4)/(2 y^2 + 1/10*2*I*y - 4)]]}, {y,  0.0001, 4}, PlotRange -> All]

I was simply explaining why the plots were different.

As for getting y as functions, respectively, of Re[x] and Im[x], that gets tricky because it is in general not a function but rather a "multivalued function". You will probably want to get a continuous branch of such inverses. One method you might consider is to find a value for y at a given re(x), say, and then trace from there by setting up an NDSolve equation.

Lets define the real and imaginary parts of f as ref and imf respectively. And the real and imaginary parts of x we will call rex and imx. The equations you have are {ref[rex,imx,y]==0,imf[rex,imx,y]==0}. To inverst y in terms of rex, treat it as a function of rex (the whole point of the exercise) and differentiate. I'll show explicit code for your example.

expr = y*Sqrt[(y^2 + 1/10*I*y - 4)/(2 y^2 + 1/10*2*I*y - 4)];
sys = {rex, imx} - ComplexExpand[{Re[expr], Im[expr]},TargetFunctions -> {Re, Im}];
dsys = D[sys /. {y -> y[rex], imx -> imx[rex]}, rex];
yinit = .1;
init = expr /. y -> yinit;

soly = NDSolveValue[Flatten[{Thread[dsys == 0], y[Re[init]] == yinit, 
    imx[Re[init]] == Im[init]}], y[rex], {rex, Re[init], 2}]

(* Out[77]= InterpolatingFunction[{{0.100125, 2.}}, <>][rex] *)

Plot[soly, {rex, Re[init], 2}]

I'm using version 10.1. You can do similarly with NDSolve if you solve for both {y[rex],imx[rex]}.

Dear Daniel, I have installed Mathematica 10, but again I can't use your code.

Attachments:

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}]

You will get better answers if you give a complete, very specific example of what you want.