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

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Faton Berisha

Posts: 39
Registered: 1/9/07
Re: complex numbers
Posted: Jan 31, 2007 7:03 PM
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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/(2|C|)).

Regards,
Faton Berisha




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