Image under a mobius transformation

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jimmycricket
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Homework Statement


Find the Mobius transformation which carries the points [tex]0,1,-i[/tex] to the points [tex]-1,0,\infty[/tex] respectively. Find the image of the domain [tex]\{z:x<0,-x+y<t\}[/tex] under this mobius transformation.

Homework Equations


The Attempt at a Solution


Let [tex]T(z)=\frac{az+b}{cz+d}[/tex].
Then [tex]T(0)=-1\Longrightarrow \frac{b}{d} \iff b=-d[/tex]

[tex]T(1)=0\Longrightarrow \frac{a+b}{c+d}=0\Longrightarrow a+b=0 \iff a=-b=d[/tex]

[tex]T(-i)=\infty\Longrightarrow \frac{-ia+b}{-ci+d}\iff d-ci=0\iff c=\frac{d}{i}=bi=-ai[/tex]

Now we have [tex]T(z)=\frac{az-a}{a-aiz}=\frac{z-1}{1-zi}[/tex]

So I now have to find the image under this map which is where I'm a bit stumped. Would it help to find where the intersections of the boundary of the domain with the axes are mapped to?
 
on Phys.org
Yes, the first part seems to be correct. :smile:
 
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Now let z= x+iy where x> 0 and x+ y< t for some fixed real number, t. For example, if t= 1, that would be the part of the complex plane to the right of the imaginary axis and below x+ y= 1. Some points on the boundary of that set would be 1, i, 2- i, etc. Determining what the transformation does to boundary points should help you see what the image is.