0
$\begingroup$
GG = {N3*Conjugate[N3]+ N4*Conjugate[N4] == 1, (2 N3 + (-I + Sqrt[2]) N4)/(2 Sqrt[3]) == 0};
Sols = Solve[GG, {N3, N4}]

I need to solve the above equation. But I didn't get an answer!!! Can you please help. This N3 and N4 will be sometimes complex number or real number or imaginary number.

$\endgroup$
1
  • 2
    $\begingroup$ FullSimplify[Sols] I get on Mathematica 12.1.1.0: {N3 -> ConditionalExpression[-(1/2) (-I + Sqrt[2]) N4, And[ Or[Im[N4] + (Rational[4, 7] - Re[N4]^2)^Rational[1, 2] == 0, Im[N4] == (Rational[4, 7] - Re[N4]^2)^Rational[1, 2]], Inequality[(-2) 7^Rational[-1, 2], LessEqual, Re[N4], LessEqual, 2 7^Rational[-1, 2]]]]} $\endgroup$
    – flinty
    Nov 19, 2020 at 0:49

3 Answers 3

2
$\begingroup$

Split the complex variables N3,N4 into real and imaginary part and ComplexExpand it.

GG = {N3*Conjugate[N3]+ N4*Conjugate[N4] == 1, (2 N3 + (-I + Sqrt[2]) N4)/(2 Sqrt[3]) == 0};

GG2 = GG /. {N3 -> n31 + I n32, N4 -> n41 + I n42}

ceRe = ComplexExpand[Re[GG2 /. Equal -> Subtract], 
         TargetFunctions -> {Re, Im}] // Simplify

ceIm = ComplexExpand[Im[GG2 /. Equal -> Subtract], 
         TargetFunctions -> {Re, Im}
] // Simplify

You get only solutions for three of the four variable, one you can choose free within a given range.

Here choose n42 from -2/Sqrt[7] to 2/Sqrt[7] and look at the solutions for the other variables.

sol = Solve[Join[Thread[ceRe == 0], {ceIm[[2]] == 0}], {n31, n32, n42, n41},
    Reals] // Simplify

Plot[Evaluate[{n31, n32, n41} /. sol[[1 ;; 2]]], {n42, -(2/Sqrt[7]), 
   2/Sqrt[7]}, PlotStyle -> {Red, Green, Blue
}]  

enter image description here

Or choose n31 free in the range +-Sqrt[3/7]

sol2 = Solve[Join[Thread[ceRe == 0], {ceIm[[2]] == 0}], {n32, n41, n31, n42},
   Reals] // Simplify

Plot[Evaluate[{n32, n41, n42} /. sol2[[1 ;; 2]]], {n31, -Sqrt[(3/7)], 
  Sqrt[3/7]}, PlotStyle -> {Green, Blue, Magenta
}]
$\endgroup$
1
$\begingroup$
Clear["Global`*"]

GG = {N3*Conjugate[N3] + N4*Conjugate[N4] == 
    1, (2 N3 + (-I + Sqrt[2]) N4)/(2 Sqrt[3]) == 0};

There are an unlimited number of solutions

Sols = Solve[GG, {N3, N4}][[1]]

enter image description here

The condition is

cond = Sols[[1, -1, -1]]

(* (Im[N4] == -(Sqrt[4 - 7 Re[N4]^2]/Sqrt[7]) && -(2/Sqrt[7]) <= Re[N4] <= 2/
    Sqrt[7]) || (Im[N4] == Sqrt[4 - 7 Re[N4]^2]/Sqrt[7] && -(2/Sqrt[7]) <= 
    Re[N4] <= 2/Sqrt[7]) *)

Some examples of N4 values

inst = FindInstance[cond, N4, 5, RandomSeeding -> 1234] // FullSimplify

(* {{N4 -> -((I (-805 I + Sqrt[2438247]))/2324)}, {N4 -> -(2/Sqrt[7])}, {N4 -> (
   I (1085 I + Sqrt[1909047]))/2324}, {N4 -> 2/Sqrt[
   7]}, {N4 -> -((I (-525 I + Sqrt[2810647]))/2324)}} *)

The associated {N3, N4} pairs are

{N3, N4} /. Sols /. inst // Simplify

(* {{((1 + I Sqrt[2]) (-805 I + Sqrt[2438247]))/
  4648, -((I (-805 I + Sqrt[2438247]))/2324)}, {(-I + Sqrt[2])/Sqrt[
  7], -(2/Sqrt[7])}, {((-I + Sqrt[2]) (1085 - I Sqrt[1909047]))/4648, (
  I (1085 I + Sqrt[1909047]))/2324}, {-((-I + Sqrt[2])/Sqrt[7]), 2/Sqrt[
  7]}, {((1 + I Sqrt[2]) (-525 I + Sqrt[2810647]))/
  4648, -((I (-525 I + Sqrt[2810647]))/2324)}} *)
$\endgroup$
1
$\begingroup$

Reduce work.

GG = {N3*Conjugate[N3] + N4*Conjugate[N4] == 
    1, (2 N3 + (-I + Sqrt[2]) N4)/(2 Sqrt[3]) == 0};
Sols = Reduce[GG, {N3, N4}]
-Sqrt[(3/7)] <= Re[N3] <= Sqrt[3/
  7] && (Im[N3] == -(Sqrt[3 - 7 Re[N3]^2]/Sqrt[7]) || 
   Im[N3] == Sqrt[3 - 7 Re[N3]^2]/Sqrt[7]) && 
 N4 == -((2 N3)/(-I + Sqrt[2]))
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.