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Argand plane graphing 
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#1
Feb2212, 10:24 PM

P: 158

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
Feb2212, 10:38 PM

HW Helper
P: 3,540

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?



#3
Feb2212, 10:48 PM

P: 158

I'm having trouble with visualizing " zz1 = k zz2" 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. 


#4
Feb2312, 12:05 AM

P: 158

Argand plane graphing
Where do z1 and z2 lie with respect to the circle that is formed? What complex number is the center?



#5
Feb2312, 01:50 AM

P: 4,573

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 (xa)^2 + (yb)^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. 


#6
Feb2312, 05:21 AM

P: 158

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)] 


#7
Feb2312, 05:55 AM

P: 158

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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