How to write this script in complex conjugate, it should have real and imaginary values

Results should have real and imaginary part in the form of (a+bi)

% Initilization 
r1source = zeros(length(y),length(x),length(z)) ; 
r2source = zeros(length(y),length(x),length(z)) ; 
r3source = zeros(length(y),length(x),length(z)) ; 

for m=1:length(y)
    for n=1:length(x)
        for o=1:length(z) 
            r1source(m,n,o)=sqrt((x(n)-x1)^2+(y(m)-y1)^2+(z(o)-z1)^2);
            r2source(m,n,o)=sqrt((x(n)-x2)^2+(y(m)-y2)^2+(z(o)-z2)^2);
            r3source(m,n,o)=sqrt((x(n)-x3)^2+(y(m)-y3)^2+(z(o)-z3)^2);
        end
    end
end

[Adam Danz]

It will return compex numbers with imaginary components only if there are negative values within the square root. Please review the concept of imaginary numbers to understand how they are created.

[Balaji Ramdas]

ok lets say from the this concept

example:

a=5
b=6
c=a+bi
real(c)
img(c)

how do i write this:

r1source(m,n,o)=sqrt((x(n)-x1)^2+(y(m)-y1)^2+(z(o)-z1)^2);

[Adam Danz]

How do the a/b/c values map to the values in the last line?

[Walter Roberson]

            r1source(m,n,o)=sqrt((x(n)-x1)^2+(y(m)-y1)^2+(z(o)-z1)^2);

if x and x1 and y and y1 and z and z1 are all real-valued, then (x(n)-x1) and (y(m)-y1) and (z(o)-z1) will be real-valued, and the square of a real quantity is never negative, so the sum of squares would never be negative, so r1source would never be complex-valued.

For r1source to be complex-valued, at least one of the quantities would have to be complex-valued. If some of the quantities are real-valued but others are purely imaginary, then although the squares of the purely imaginary components would be negative, they might not be negative enough to balance the other parts, so you could end up with sqrt() of a positive number.

Answer 1

[Walter Roberson]

r1source = sqrt( (reshape(y,[],1)-y1).^2) + (reshape(x,1,[])-x1).^2 + (reshape(z,1,1,[])-z1).^2 );