How to solve this set of equation in complex domain?

Question

First, I got a real version of equation set

Solve[{d u^2 - d v^2 == 2 e u v, (u u + v v) == 1}, {u, v}]

This can be correctly solved in the blink of an eye.

Now I want to solve this complex version

Solve[{d u^2 - d\[Conjugate] v^2 - 2 e u v == 
   0, (u u\[Conjugate] + v v\[Conjugate]) == 1}, {u, v}]

mathematica can not be solve the above code.

But I can show how this set of equation can be easily solved.

take

$$ - {v^2}{d^*} + d{u^2} - 2euv = 0$$

multiply both side by $\frac{{{d^*}}}{{{u^2}}}$, we got $$- {\left( {{d^*}\frac{v}{u}} \right)^2} - 2e\left( {{d^*}\frac{v}{u}} \right) + {\left| d \right|^2} = 0$$ we can see this is a simple quadratic equation relative to ${d^*}\frac{v}{u}$, and got $${d^*}\frac{v}{u} = e \pm \sqrt {{e^2} + {{\left| d \right|}^2}} $$ define $$\xi \equiv \sqrt {{e^2} + {{\left| d \right|}^2}} $$ then $$v = \frac{u}{{{d^*}}}\left( {e + \xi } \right)$$ and $${\left| v \right|^2} = \frac{{{{\left| u \right|}^2}}}{{{{\left| d \right|}^2}}}{\left( {e + \xi } \right)^2}$$ plug this back into $${\left| u \right|^2} + {\left| v \right|^2} = 1$$ we got $${\left| u \right|^2}\left[ {1 + \frac{1}{{{{\left| d \right|}^2}}}{{\left( {e + \xi } \right)}^2}} \right] = 1$$ at this step, we can say we solved the absolute value of u and v. their relative phase is determined by previous $v = \frac{u}{{{d^*}}}\left( {e + \xi } \right)$

After I solved the equation, I notice that I missed a condition in Solve. Because there is phase uncertainty, if we restrict that u is real, then v should be also fixed.

So I thought this can be solved

Solve[{d u^2 - d\[Conjugate] v^2 - 2 e u v == 
   0, (u u\[Conjugate] + v v\[Conjugate]) == 1, Im[u]==0}, {u, v}]

With Im[u] condition added.

But mathematica still can not give an answer. Am I missing something here? How to solve this set of equation?

Answer 1

Supose, only v and d are complex, ComplexExpand gives a fast result:

ce = ComplexExpand[{d u^2 - d\[Conjugate] v^2 - 2 e u v == 
       0, (u u\[Conjugate] + v v\[Conjugate]) == 1}, {v, d}, 
       TargetFunctions -> {Re, Im}] /. {Re[d] -> d1, Im[d] -> d2, 
       Re[v] -> v1, Im[v] -> v2}

(*   {d1 u^2 - 2 e u v1 - d1 v1^2 - 2 d2 v1 v2 + d1 v2^2 + 
      I (d2 u^2 + d2 v1^2 - 2 e u v2 - 2 d1 v1 v2 - d2 v2^2) == 0, 
      u^2 + v1^2 + v2^2 == 1}   *)

Both, real and imaginary part have to be zero.

Solve[{d1 u^2 - 2 e u v1 - d1 v1^2 - 2 d2 v1 v2 + d1 v2^2 == 
       0, (d2 u^2 + d2 v1^2 - 2 e u v2 - 2 d1 v1 v2 - d2 v2^2) == 0, 
       u^2 + v1^2 + v2^2 == 1}, {u, v1, v2}] // Simplify

Answer 2

Substituting

u = \[Rho]u Exp[I \[Alpha]u];
v = \[Rho]v Exp[I \[Alpha]v];
d = \[Rho]d Exp[I \[Alpha]d];
gln = Simplify[{d u^2 - d\[Conjugate] v^2 - 2 e u v ==0, (u u\[Conjugate] + v v\[Conjugate]) == 1},Assumptions ->Element[{\[Rho]u, \[Rho]v, \[Rho]d, \[Alpha]u,\[Alpha]v, \\[Alpha]d}, Reals]]

you can solve directly

sol = FullSimplify[Solve[gln, {\[Rho]u, \[Rho]v(*,\[Alpha]u,\[Alpha]v*)}],Assumptions ->Element[{\[Rho]u, \[Rho]v, \[Rho]d, \[Alpha]u, \[Alpha]v, \\[Alpha]d}, Reals]]