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The question is what is the method to solve the implicit function has real and imaginary number.

For example, The function is $$F(x,y)=(-I*x + 2*y^2)^2 + x^2 - 4*y^4*Sqrt[1 - I*x/(y^2)]$$

Although I used a NSolve and Solve I can't get the solution. (It is too complicated to solve by those command i think.)

What is the value y? (y may be a complex number when x is given) And is it possible to draw a graph of Real[y] - x and Imaginary[y] - x?

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Have a look at FindRoot. –  b.gatessucks Feb 26 '13 at 8:02
    
Do you want a solution for the equation F[x,y]==c? –  Sjoerd C. de Vries Feb 26 '13 at 9:26
    
yes. (c==0) Do you know how to draw the graph about that relation? such as Real[y]-x (x is a real number. y is a complex number) –  user132682 Feb 26 '13 at 10:12
1  
@user6108 Could you explain why don't you upvote and accept answers to most of your questions? –  Artes Dec 16 '13 at 0:45

1 Answer 1

Solve works, but first let's clean up the definition to avoid unnecessary singularities:

f[x_, y_] := (-I*x + 2*y^2)^2 + x^2 - 4*y^4*Sqrt[1 - I*x/(y^2)];
f[x, y] // TraditionalForm

$ x^2+\left(2 y^2-i x\right)^2-4 y^4 \sqrt{1-\frac{i x}{y^2}}$

This is easily simplified by inspection, but we can get Mathematica to help us. I cheat a little by giving it some overly strong assumptions; the purpose is to get it to "recognize" that the $1/y^2$ term is a removable singularity:

Assuming[y > 0, FullSimplify[f[x, y]]] // TraditionalForm

$4 y^2 \left(y \left(y-\sqrt{y^2-i x}\right)-i x\right)$

OK, that looks good, so let's memorialize it:

g[x_, y_] := Evaluate[Assuming[y > 0, FullSimplify[f[x, y]]]]

Now we invoke Solve:

Solve[g[x, y] == 0, {y}]

$\left\{\{y\to 0\},\left\{y\to -\sqrt[4]{-1} \sqrt{x}\right\},\left\{y\to \sqrt[4]{-1} \sqrt{x}\right\}\right\}$

(It is wise to pause here to reflect on the nature of these solutions due to ambiguities in the square and fourth roots: there are two square roots and four fourth roots involved in each solution, potentially designating $16$ different functions in all. However, it is evident that if we allow $x$ to be negative and permit complex square roots to be computed, then we need consider only the primitive fourth root $\exp{i \pi / 4}$ and its negative, both of which show up in the last pair of solutions. We conclude that we should retain all the solutions Solve has provided and allow $x$ to range over all real values, positive and negative.)

Let's encapsulate these solutions within yet another function. This re-solves the equations and extracts the solutions as expressions:

h[x_] := Evaluate@(Flatten[Solve[g[x, y] == 0, {y}]] /. Rule[y, f_] :> f)
h[x] // TraditionalForm

$\left\{0,-\sqrt[4]{-1} \sqrt{x},\sqrt[4]{-1} \sqrt{x}\right\}$

To plot a complex function of a real argument $x$, $y(x)$, we can employ three dimensions: its graph will be a curve. Because there are three separate solutions, we get parts of three such curves using ParametricPlot3D. Its three components are the parameter $x$ and the real and imaginary parts of $y(x)$. I give a fully general expression by selecting random colors for the curves: this avoids even having to count the number of distinct solutions in advance. So that the complex part is appropriately rendered, I use the BoxRatios argument to specify that the last two dimensions (real and imaginary axes) have the same scales.

ParametricPlot3D[Evaluate[{x, Re[#], Im[#]} & /@ h[x]], {x, -2, 2}, 
 PlotStyle -> Table[{Thick, Hue[RandomReal[], .8, .8]}, {i, h[0]}], 
 BoxRatios -> {Automatic, 1, 1}, Boxed -> False, 
 AxesLabel -> {x, Re[y], Im[y]}]

Plot 3D

You can also use ParametricPlot to render the image of $y$ in the complex plane, perhaps distinguishing the values of $x$ by color. Just drop the first parameter, leaving the real and imaginary parts to be drawn:

ParametricPlot[Evaluate[{Re[#], Im[#]} & /@ h[x]], {x, -2, 2}, 
 PlotStyle -> Thickness[0.02],  
 ColorFunction -> Function[{t, x, y}, Hue[t]], 
 AxesLabel -> {Re[y], Im[y]}, LabelStyle -> Medium]

Plot 2D

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