# Image of the lines y=x and y=-x under a Mobius Transformation

Question: Consider the Mobius transformation

$$w=\frac{z-1}{z-i}$$

Find the image under this transformation of the unit circle and the lines $$y=\pm x$$, where $$z=x+iy$$.

This is an example out of a textbook that I just can't wrap my head around. According to the answers, the image of the unit circle is the line $$x=-y$$, the line $$y=x$$ is mapped to the unit circle and the line $$y=-x$$ is mapped to the unit circle with centre $$-1+i$$.

The first one with the unit circle I figured out by substituting points on the unit circle into $$w$$. However, using the same method for the the line $$y=x$$, I found that every point lying on the line was mapped to $$\frac{-1}{-i}=i$$ so I don't see how that can become the unit circle. With substituting points lying on the line $$y=-x$$ I got a whole bunch of points which seem like they lie on the circle of radius $$1$$ with centre $$1-i$$ rather than $$-1+i$$.

What have I done wrong? I used $$z=x+iy$$ to get $$w=\frac{x+iy-1}{x+iy-i}$$ and substituted in the appropriate $$(x,y)$$ coordinates.

• When $y=x$ you get $w=\frac {x-1+ix} {x+i(x-1)}$. This is not equal to $i$. Nov 15, 2019 at 10:17

The three things that I always try to bear in mind when working with Möbius transformations are:

1. Consider them to be defined on $$\hat{\mathbb{C}}$$, so that you also take into account the point at infinity ($$\infty$$). Then a line is just a circle that contains the point $$\infty$$.
2. Circles (and hence lines) are mapped into circles (which might be lines).
3. Any circle (or line) is determined by three points (if it is a line you already know that one of these three points is $$\infty$$).

So to know the image of a circle (or line) you only need to find the images of three of its points and see what circle (or line) they determine. Let's do this in your case:

• Three points in the unit circle are $$1$$, $$i$$, $$-i$$. Under your transformation they are mapped to $$0$$, $$\infty$$, $$\frac{1-i}{2}$$, which are all on the line $$y=-x$$.
• Three points on the line $$y=x$$ are $$0$$, $$1+i$$, $$\infty$$. They are mapped to $$-i$$, $$i$$, $$1$$, which are all on the unit circle.
• Three points on the line $$y=-x$$ are $$0$$, $$1-i$$, $$\infty$$. They are mapped to $$-i$$, $$\frac{2-i}{5}$$, $$1$$, which all lie on the unit circle with center at $$1-i$$ (you are right on that).
• Thanks for the explanation. For $y=-x$, how would I determine that the mapped points lie on the unit circle centred at $1-i$ if I didn't already know this beforehand? Nov 15, 2019 at 11:28
• The center of the circle is the circumcentre of the triangle with vertices on $-i,\frac{2-i}{5}, 1$. Nov 16, 2019 at 18:16