Question: Consider the Mobius transformation


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.

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

1 Answer 1


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).
  • $\begingroup$ 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? $\endgroup$
    – Xtrfyable
    Nov 15, 2019 at 11:28
  • $\begingroup$ The center of the circle is the circumcentre of the triangle with vertices on $-i,\frac{2-i}{5}, 1$. $\endgroup$ Nov 16, 2019 at 18:16

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