Turning imaginary numbers into real numbers

Hello there!

I am attempting to find the points where n = S with the following equations. Whenever S goes into the imaginary plane though, I want it to convert to S = 1. Even though I have a check for that in this code, for some reason MATLAB seems to be ignoring that bit and converting S into 0 instead. How can I fix this?

 for k = 1:1:100
    n = (100 / k);
    s =  sqrt((((.25*(n-(.5*(100-n))))/(n))-.25)/(-((n - .5*(100-n)))/(n)));
    if (0<=s) && (s<=1) 
        s = s;
    elseif s > 1 
        s = 1;
    elseif (isnan(s)==1)
        s = 1;
    elseif (isreal(s)==0)
        s = 1;
    else 
        s = 1;
    end
    RTDlineark{k} = [k]; 
    RTDlinearkn{k} = [n];
    RTDlinearks{k} = [s];
end

Answer 1

[Birdman]

To get rid of that, your code can be improved as follows:

   for k = 1:1:100
      n = (100 / k);
      s =  sqrt((((.25*(n-(.5*(100-n))))/(n))-.25)/(-((n - .5*(100-n)))/(n)));
      if (0<s) && (s<=1) 
          s = s;
      elseif s > 1 
          s = 1;
      elseif (real(s)>=0) && imag(s)~=0
          s = 1;
      end
      RTDlineark{k} = [k]; 
      RTDlinearkn{k} = [n];
      RTDlinearks{k} = [s];
  end

[Walter Roberson]

This happens to work with the data, but still suffers from the problem of using < and <= operators on complex quantities, as those operators are defined to ignore the complex component.

Answer 2

[Andrei Bobrov]

k = 1:100;
n = 100 ./ k;
s = sqrt( (.25*(n-.5*(100-n))./n-.25)./-(n - .5*(100-n))./n );
t = abs(imag(s)) > 0;
s(t) = t(t) + 0;

Answer 3

[Walter Roberson]

The relative comparison operations ignore the imaginary component. You need to do the test for imaginary first.