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Topic:
complex numbers
Replies:
8
Last Post:
Jan 31, 2007 7:33 PM




Re: complex numbers
Posted:
Jan 31, 2007 7:03 PM


On Jan 31, 7:52 pm, "Stephen J. Herschkorn" <sjhersc...@netscape.net> wrote: > jennifer wrote: > [snip] > >Also how do i describe a family of curves in the complex plane: > > >1. Re(1/z)= C > > >The answer is a family of circles centered at 1/(2C)+0i of radius 1/ > >modulus 2C but with the origin excluded. > > That should be 1 / (2 C). > > >My question is how did they get the center as 1/(2C)+0i and radius 1/ > >modulus 2C. > > >I got the Re(1/z)= x/(x^2+y^2) = C but how do i obtain the radius and > >the center as indicated above? > > So x^2  x / C + y^2 = 0. Complete the square. > > >I know the equation of a circle is: x^2 +y^2= r^2 > > That is for a circle centerer at the origin. Do you know the equation > for an arbitrary center in the plane? >
Or if you like it better, without switching to coordinates,
C = Re(1/z) = 1/2 (1/z + 1/ bar z) = 1/2 (z + bar z)/(z bar z);
hence,
z + bar z = 2Cz bar z,
or
(z  1/(2C)) (bar z  1/(2C)) = (1/(2C))^2.
Since 1/(2C) is real,
(z  1/(2C)) bar (z  1/(2C)) = (1/(2C))^2,
hence
z  1/(2C)^2 = (1/(2C))^2,
or
z  1/(2C) = (1/(2C)).
Regards, Faton Berisha



