# Trouble in mapping of möbius transformation

Question:-

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

• The circle is centered at $(2,0)$ though. – Chris Custer May 19 at 15:54
• I apologize for that mistake. Although i have corrected that but still the problem is same – Vedant Chourey May 19 at 16:02
• Yes. Perhaps you can take three points (Möbius transformations are determined by their effect on three points). – Chris Custer May 19 at 16:05
• Nice points, like $(0,0),(4,0)$ and $(2,2)$. – Chris Custer May 19 at 16:07
• I will try this method – 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 $$4w+3=0$$

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