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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.

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  • $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Apr 14 at 23:25
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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.

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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 $(1,0)$.

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  • $\begingroup$ 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? $\endgroup$ – BesMath Apr 15 at 0:26
  • $\begingroup$ @BesMath - it’d be intuitive to understand a map, but not rigorous enough to determine the map $\endgroup$ – Quanto Apr 15 at 0:33

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