Show that the transformation $$ w = \frac{2z+3}{z-4}$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$

My attempt:- The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$ So the inverse mapping of the given bilinear transformation is:- $$z= \frac{4w+3}{w-2} $$ Now substituting the value of $z$ in $(1)$ $$\frac{|3w+1|}{|w-2|} =2$$ $$|4w+3|=2|w-2|$$ $$|3u+2+3v\iota|=2|u-2+v\iota|$$ $$9u^2+4+12u+v^2= 4u^2+16-16u+v^2$$ On solving these it appears as $$5u^2+28u-12=0$$

I can not come at the conclusion as stated in question, is my method correct ?

Suggestions are highly appreciated Thankyou

  • $\begingroup$ The circle is centered at $(2,0)$ though. $\endgroup$ – Chris Custer May 19 at 15:54
  • $\begingroup$ I apologize for that mistake. Although i have corrected that but still the problem is same $\endgroup$ – Vedant Chourey May 19 at 16:02
  • $\begingroup$ Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points). $\endgroup$ – Chris Custer May 19 at 16:05
  • $\begingroup$ Nice points, like $(0,0),(4,0)$ and $(2,2)$. $\endgroup$ – Chris Custer May 19 at 16:07
  • $\begingroup$ I will try this method $\endgroup$ – Vedant Chourey May 19 at 16:12

If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get

$$\Big|{|4w+3 -2w+4\over w-2}\Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$

so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-{3\over 4}$ or $$4w+3=0$$

  • $\begingroup$ Thankyou. For explanation $\endgroup$ – Vedant Chourey May 19 at 16:12
  • $\begingroup$ Fell free to upvote and accept the answer if you think it was usefull answer to you. $\endgroup$ – Aqua May 19 at 16:14
  • $\begingroup$ Ok. I'll take care of this next time $\endgroup$ – Vedant Chourey May 19 at 16:26

$0\to-\dfrac 34, 4\to\infty $ and $2+2i\to -\dfrac 34-\dfrac{11}4i$.

The result follows.


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