Image under a mobius transformation

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SUMMARY

The Mobius transformation that maps the points 0, 1, and -i to -1, 0, and ∞ respectively is given by T(z) = (z - 1) / (1 - zi). This transformation was derived by setting up equations based on the mappings of the specified points. The next step involves determining the image of the domain defined by {z: x < 0, -x + y < t} under this transformation, which can be approached by analyzing how the boundary points of the domain are transformed.

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Homework Statement


Find the Mobius transformation which carries the points 0,1,-i to the points -1,0,\infty respectively. Find the image of the domain \{z:x&lt;0,-x+y&lt;t\} under this mobius transformation.

Homework Equations


The Attempt at a Solution


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

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

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

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

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?
 
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Yes, the first part seems to be correct. :smile:
 
Last edited:
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.
 

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