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 Exercise 8.4 
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Joined: 04 Feb 2009, 15:52
Posts: 8
Post Exercise 8.4
Regards to you all.

I'm beginner, but I enjoy discussing. And I dare to discuss about the solution of exercise 08.04 written by Sameed Zahoor.

I think the equation z\overline{z}=k is only good at circles centered at the origin. In general, the equation of a circle must be (z-z_0)\overline{(z-z_0)}=k where k is constant (the radius squared) and z_0 is the certer of the circle.

If I'm right about this, the answer to the problem is not as easy as it looks.

Here is my solution (too long for my taste, but I wish it's right):

Suppose a circle \Gamma of raious R and center at point C=s.e^{i\alpha}. I'll use the equation of circle in polar cordinates (this equation cames from the Law of cosines). So, let's


First of all, I need to trace a straight line from origin to the center of the circle and extend this line infinitly in both directions. This line will cut the circle at two points. I'll name those points A and B, and the center of the circle I named C,
08_04_d1.JPG [ 10.18 KiB | Viewed 1566 times ]

C=s.e^{i\alpha} where s is the distance between C and the origin, and \alpha is the angle of the line. I'll study the first case, s>R. Then,






Now, I say that \Gamma'=f(\Gamma) is a circle with center C' and radious \frac{R}{s^2-R^2} where C'=\frac{1}{2}(A'+B'),


If I write C'=s'.e^{-i\alpha} \Rightarrow s'=\frac{s}{s^2-R^2} and note that the center C' of \Gamma' is not f(C).

I can say this, because

\forall X\in\Gamma, X=r.e^{i\theta}, f(X)=\frac{1}{r}.e^{-i\theta}

and I'll prove that f(X) obey a equation of a circle, just like this:

(\frac{1}{r})^2-2.\frac{1}{r}.s'.cos(-\theta-(-\alpha))+s'^2 =


And owing to the definition of \Gamma, then 2s.cos(\theta-\alpha)=\frac{r^2+s^2-R^2}{r}, and the previous expression turns into




So f(X)=\frac{1}{r}.e^{-i\theta} is in a circumference of radius \frac{R}{s^2-R^2} and center \frac{s}{s^2-R^2}.e^{-i\alpha}


The second case, R>s, (the origin is inside the circle) gives the points A,B like this:

B=(R+s)e^{i\alpha}; A=(R-s)e^{i(\alpha+\pi)}


B'=f(B)=\frac{1}{R+s}.e^{-i\alpha} A'=f(A)=\frac{1}{R-s}.e^{-i(\alpha+\pi)}=\frac{1}{R-s}.e^{-i\alpha}.e^{-i\pi}=


The expressions of A' and B' are the same of those expressions of the first case, so all the demostration is the same.

All I have to do is study the third case, R=s (I think \Gamma' turn to straight line)...

I was looking for a solution for a week, and this is only what I could done. I'm sure there must be a better and shorter proof. Please, would you answer this message?

See you soon!!

Last edited by Pibe on 16 Feb 2009, 21:21, edited 1 time in total.

12 Feb 2009, 18:23

Joined: 30 Jun 2008, 22:14
Posts: 25
Post Re: Exercise 8.4
You're absolutely right. Circle at origin is not a general case. A shorter proof probably involves a Mobius transformation, in this case a translation, that transforms a general circle to one at the origin.

12 Feb 2009, 20:27

Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise 8.4
I have put a short general solution to this problem in the solutions section under Exercise [08.04] b.
Let me know what you think.

15 Feb 2009, 16:57
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