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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?

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  • $\begingroup$ 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
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$\frac{1+z}{1-z}$ 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 $w=i \cot \frac{\alpha}{2}$

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  • $\begingroup$ but the answer in my book is : The circle u^2 +v^2-2u*ctgα=1... $\endgroup$ – needtostudy Aug 4 '13 at 9:36
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    $\begingroup$ @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

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