Exercise 8.4

Pibe · 12 Feb 2009, 18:23

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

$\Gamma=\{r.e^{i\theta}|r^2-2sr.cos(\theta-\alpha)+s^2=R^2\}$

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,

Attachment:

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,

$A=(s-R).e^{i\alpha}$

$B=(s+R).e^{i\alpha}$

So,

$A'=f(A)=\frac{1}{s-R}e^{-i\alpha}$

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

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

$C'=\frac{1}{2}(\frac{1}{s-R}+\frac{1}{s+R}).e^{-i\alpha}=\frac{s}{s^2-R^2}.e^{-i\alpha}$

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:

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

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

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

$\frac{1}{r^2}-\frac{(r^2+s^2-R^2)}{r^2(s^2-R^2)}+\frac{s^2}{(s^2-R^2)^2}=$

$=\frac{(s^2-R^2)^2-(r^2+s^2-R^2)(s^2-R^2)+s^2r^2}{r^2(s^2-R^2)^2}=\frac{s^4-2s^2R^2+R^4-(r^2s^2-r^2R^2+s^4-s^2R^2-R^2s^2+R^4)+s^2r^2}{r^2(s^2-R^2)^2}=$

$=\frac{r^2R^2}{r^2(s^2-R^2)^2}=\frac{R^2}{(s^2-R^2)^2}=const.$

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

...uf

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

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

and

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

$=\frac{-1}{R-s}.e^{-i\alpha}=\frac{1}{s-R}.e^{-i\alpha}$

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!!