# I need help understand this Möbius transformation

Show $$w=\frac{z-i}{z+i}$$ maps upper half plane into a unit disk centered at origin.

I rewrote the equation as $$z=-i(\frac{w+1}{w-1})$$ and since $$|z|>0$$ on upper half plane. I say $$|-i(\frac{w+1}{w-1})|>0$$ which is $$|w+1|>|w-1|$$ so $$|(u+1)+iv|>|(u-1)+iv|$$ I am stuck then. because this will not end up with a circle formula.
I had seen examples how a function maps circles to half plane, using three points. but here I need some help to understand how to work on this transformation. Thank you.

• For complex $z$, $|z|\gt 0$ everywhere but the origin. Your characterization of the upper half-plane is wrong; you need to be a little more careful about manipulating absolute values with complex numbers. – Steven Stadnicki Apr 14 at 23:25

This is the Cayley transformation. Note that $$1\mapsto-i,0\mapsto-1, \infty\mapsto1$$.

Mobius transformations map generalized circles to generalized circles, and are completely determined by the images of three points.

Thus a test point, $$i\mapsto0$$ and we are done.

Let $$Im(z) = y \in [0,\infty)$$ for the upper plane and rewrite $$w=\frac{z-i}{z+i}$$ as $$z=\frac{1+w}{1-w}i$$. Then,

$$z-\bar z = 2iy = \frac{1+w}{1-w}i + \frac{1+\bar w}{1-\bar w}i$$

Rearrange to get $$(1+y)|w|^2 - y(w+\bar w) = 1-y$$. Put it in the form of complex circles,

$$|w-\frac y{1+y}|^2 = \frac1{(1+y)^2}$$

which represents a unit disk centered at origin. In particular, for $$y=0$$, it has its center at $$(0,0)$$ and radius 1; and for $$y\to\infty$$, it converges to the point $$(1,0)$$.

• why can t we write this as $w=1-\frac{2i}{z+i}$ ? and apply the mappings step by step. I tried this and I could not reach to a unit circle. but I do not see why not? – BesMath Apr 15 at 0:26
• @BesMath - it’d be intuitive to understand a map, but not rigorous enough to determine the map – Quanto Apr 15 at 0:33