So I am giving this important exam on complex analysis on September the 12th and I'm preparing for it.I found this exercise in a book: In what lines of the plan $C_w$ are the mapped:

a)The ray $\operatorname{arg} z$=$α$ using the function $w= (1+z)/(1-z)$

b) The circles $ |z|=r$, where 0< r<1 using the function $w= 0.5\cdot [ z+ (1/z)]$

So can you please give me a clue,because I don't even know where to start...

Edit: So I figured out now that the second is an eclipse,but how about the first one?

  • $\begingroup$ Do you mean the line $\arg z = \alpha$, instead of radius? $\endgroup$ – user61527 Aug 4 '13 at 9:01
  • $\begingroup$ No in my book it says radius... $\endgroup$ – needtostudy Aug 4 '13 at 9:02
  • $\begingroup$ The set of points with a given argument is a ray / line segment, while the set of points with a given modulus is a circle. $\endgroup$ – user61527 Aug 4 '13 at 9:08
  • $\begingroup$ ah its ray,im sorry :) $\endgroup$ – needtostudy Aug 4 '13 at 9:10

$\frac{1+z}{1-z}$ is a Möbius transformation, hence it maps generalized circles onto generalized circles. (a generalized circle is a circle or a straight line). The closed ray $\arg z=\alpha$ contains the points $z=0,\infty,e^{i \alpha}$ with images $w=0,-1,\frac{1+e^{i \alpha}}{1-e^{i \alpha}}=0,-1,i \cot \frac{\alpha}{2}$.

Thus the image of the ray $\arg z=\alpha$ is the circular arc (or straight) from $w=0$ to $w=-1$ through the point $w=i \cot \frac{\alpha}{2}$

  • $\begingroup$ but the answer in my book is : The circle u^2 +v^2-2u*ctgα=1... $\endgroup$ – needtostudy Aug 4 '13 at 9:36
  • 1
    $\begingroup$ @Mathfreak you know three points on the circular arc, from that you can deduce the center and the radius. $\endgroup$ – user1337 Aug 4 '13 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.