I want to find the value of complex E_k and F_k

clear all
clc

x = -0.5:0.0294:0.5;
y = -0.5:0.0294:0.5;

syms E_k F_k
a=0.005;
z = x + 1i * y;
s1=0;

for k = 1 : 15
    A = E_k * sqrt(z.^(2)-a^(2) * z.^(2*k-2)) + sqrt((conj(z).^2) - a.^2 * conj(z).^(2*k-2));
    B = F_k * (z.^(2*k-1) - (conj(z).^(2*k-1)));
    C = E_k * ((2*k-1) * z.^(2*k-1) - (2*k-2) * a.^2 * z.^(2*k-3));
    D = F_k * ((2*k-1) * z.^(2*k-2));
end

C = rdivide(C, sqrt(z.^2 - a.^2));
E = C + D;
E = conj(E);
E = (z - conj(z)) .* E;

F = -(2.5 * sin(45) - cos(45)) * z + 1i * (sin(45) + 2.5*cos(45)) * conj(z);

lhs = A + B + E + F;


plhs = real(lhs);
qlhs = imag(lhs);


rhs_x = 0;
for k1 = x
    rhs_x = rhs_x + 2.5 * k1;
end

rhs_y = 0;
for k1 = y
    rhs_y = rhs_y + 1i * k1;
end

rhs = 1i * (rhs_x + rhs_y);

prhs = real(rhs);
qrhs = imag(rhs);

syms E_k F_k
eqn = [plhs == prhs, qlhs == qrhs];
[E_k, F_k] = solve(eqn, E_k, F_k);

[Paul]

As best I can tell, eqn can be written in the form A*x = b, where x = [E_k; F_k] and A is a 70 x 2 matrix. Is there good reason to think that such an overdetermined system will have a solution?

Also, the trig functions cos and others take input argument in radians. Use cosd and others for input argument in degrees.

[sudhanshu rana]

How can I write(E_K,F_K) in term of x? As we know that E_K and F_K are the two complex variables that is unknown.so that equation are like that only. I attached the photo of the equation. Can we please help me out to find the value of E_K,F_K.

Answer 1

[Kartik]

Hi,

To express the values of E_k and F_k in terms of x, you need to replace z with x + 1i*y in your equations and solve for E_k and F_k using the 'solve' function. Here is the modified code:

clear all
clc

x = -0.5:0.0294:0.5;
y = -0.5:0.0294:0.5;

syms E_k F_k
a=0.005;
s1=0;

for k = 1 : 15
    z = x + 1i * y;
    A = E_k * sqrt(z.^(2)-a^(2) * z.^(2*k-2)) + sqrt((conj(z).^2) - a.^2 * conj(z).^(2*k-2));
    B = F_k * (z.^(2*k-1) - (conj(z).^(2*k-1)));
    C = E_k * ((2*k-1) * z.^(2*k-1) - (2*k-2) * a.^2 * z.^(2*k-3));
    D = F_k * ((2*k-1) * z.^(2*k-2));
    C = rdivide(C, sqrt(z.^2 - a.^2));
    E = C + D;
    E = conj(E);
    E = (z - conj(z)) .* E;
    F = -(2.5 * sin(45) - cos(45)) * z + 1i * (sin(45) + 2.5*cos(45)) * conj(z);
    lhs = A + B + E + F;
    plhs = real(lhs);
    qlhs = imag(lhs);
    rhs_x = 0;
    for k1 = x
        rhs_x = rhs_x + 2.5 * k1;
    end
    rhs_y = 0;
    for k1 = y
        rhs_y = rhs_y + 1i * k1;
    end
    rhs = 1i * (rhs_x + rhs_y);
    prhs = real(rhs);
    qrhs = imag(rhs);
    eqn = [plhs == prhs, qlhs == qrhs];
    [solE, solF] = solve(eqn, E_k, F_k);
    E_k_sol(:,k) = double(subs(solE)); % Solution of E_k for kth iteration
    F_k_sol(:,k) = double(subs(solF)); % Solution of F_k for kth iteration
end

The modified code uses a loop to solve for E_k and F_k for each value of x. The solutions are stored in the arrays E_k_sol and F_k_sol, where each column corresponds to a different value of k.