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Argand plane graphing

  1. Feb 22, 2012 #1
    I'm having trouble with the various geometrical representation of complex numbers.
    Can someone provide me link where this is discussed, or maybe an argand plane graphing calculator online?
  2. jcsd
  3. Feb 22, 2012 #2


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    Homework Helper

    Is there anything in particular that's troubling you or is it a general problem you're having? Are you studying from a textbook or in school?
  4. Feb 22, 2012 #3
    I'm having trouble with visualizing " |z-z1| = k |z-z2|" where k is an real number. I know that it gives a circle, but where are the complex numbers z1 , z2 located on this circle? What deterrence do the different values of K make? (less than 0, more than 0, less than 1,etc.)

    That is just one problem I came across in my book, but I'd love to see all the other different geometrical shapes and representations in the argand plane.
  5. Feb 23, 2012 #4
    Where do z1 and z2 lie with respect to the circle that is formed? What complex number is the center?
  6. Feb 23, 2012 #5


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    Hey AlchemistK.

    Try expanding the equation until you get the equation for a circle.

    Consider that z = a + ib. What you want to do is get an equation in terms of the x and y coordinates, but in terms of an ellipse or circle: a circle has the equation (x-a)^2 + (y-b)^2 = r^2 for a circle centred at (a,b) with a radius r.

    So arrange your equation in terms of your z1 and z2 by making them constant (z1 = c + di, z2 = e + fi) and solve in terms of your z (make z = a + bi). You're a and b terms will be variable (like say x and y in a normal cartesian function) and the c,d,e,f terms are just constants.

    Rearrange them so that you an equation like (t - a)^2 + (u - b)^2 = r^2 for some constants t,u, and r where a and b correspond to the z = a + bi representation of z.

    If you do this you will understand all the concepts and it will help you with later mathematics.
  7. Feb 23, 2012 #6
    OK on taking z=x+iy, z1 = a+ib and z2= c+id and solving I get :

    x^2 + y^2 + 2x[(a - c*k^2)/(k^2 -1)] + 2y[(b - d*k^2 )/(k^2 -1)] - R

    Where R is some constant. So the center comes out to be :

    [(c*k^2 -a)/(k^2 -1)] , [(d*k^2 -b)/(k^2 -1)]
  8. Feb 23, 2012 #7
    For K greater than 1, the center lies closer to z2 and the other way around for z1.

    So I can compare which one lies closer and hence should be in the interior, but I cant calculate the radius (very complex to solve) so I cant tell if there will be cases when both z1 and z2 are inside/outside/on the circle.
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