# Image under a mobius transformation

1. Mar 25, 2014

### jimmycricket

1. The problem statement, all variables and given/known data
Find the Mobius transformation which carries the points $$0,1,-i$$ to the points $$-1,0,\infty$$ respectively. Find the image of the domain $$\{z:x<0,-x+y<t\}$$ under this mobius transformation.

2. Relevant equations

3. 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?

2. Mar 26, 2014

### z.js

Yes, the first part seems to be correct.

Last edited: Mar 26, 2014
3. Mar 26, 2014

### HallsofIvy

Staff Emeritus
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.