# Show that Mobius transformation $f(z) = \frac{z-i}{z+i}$ maps upper-half plane to a unit disk

Question: Show that the mobius transformation $$f(z) = \frac{z-i}{z+i}$$ maps the upper half plane to the unit disk.

My attempt:

Consider the imaginary axis $$iy$$ where $$0 \le y \le 1$$. $$f(z) = \frac{y-1}{y+1}$$. Easy to see that $$-1 \le f(z) \le 0$$.

Consider the imaginary axis $$iy$$ where $$1 \le y < \infty$$. $$f(z) = \frac{y-1}{y+1}$$. Easy to see that $$0 \le f(z) \le +1$$.

If $$z$$ is real, then $$|f(z)| = |\frac{z-i}{z+i}| = 1$$, so the real axis gets mapped to the unit circle.

Now I can't find an elegant way to prove that the rest of the points in the upper half plane get mapped to the interior of the unit circle.

Any hint would be appreciated. Thanks

• I just thought of something. Since $|\frac{a}{b}| = \frac{|a|}{|b|}$, we get $|f(z)| = |\frac{z-i}{z+i}| = \frac{|x + i(y-1)|}{|x + i(y+1)|} = \frac{x^2 + (y-1)^2}{x^2 + (y+1)^2} \le 1$ assuming $y >= 0$. This should be a good enough proof, right? – sku Mar 28 '20 at 23:16
• Hint: If $z\in\Bbb R$, then $|f(z)|=1$, so the real line is mapped to the unit circle, and $f(i)=0$ so the upper half plane goes inside the circle and the lower half plane goes outside. – Berci Mar 28 '20 at 23:20
• Dear @Berci, I don't understand how your conclusion is derived from the two facts that real axis mapped to unit circle and $f(i) = 0$. Is there a theorem or something to this effect? Sorry, I am still learning. – sku Mar 28 '20 at 23:28

Let $$w= \frac{z-i}{z+i}$$. Then, $$z = i\frac{1+w}{1-w}$$. Denote the upper half of the $$z$$-plane as $$Im(z) = y \ge 0$$. Then,

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

Rearrange to get $$(1+y)|w|^2 - y(w+\bar w)=1-y$$, or in the explicit equation of a circle,

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

which shows that each horizontal line of $$y\in[0,\infty)$$ in the upper plane maps onto a circle of the center $$\frac{y}{1+y}$$ and the radius $$\frac1{1+y}$$, as shown in the graph, As seen in the graph, the real axis $$y = 0$$ maps to the unit circle $$|w| =1$$. As $$y$$ increases, the center moves towards $$w=1$$ and the radius decreses, eventually converging to the point $$1$$ as $$y\to \infty$$. Thus, $$w= \frac{z-i}{z+i}$$ maps the upper half of the $$z$$-plane onto the unit disk $$|w|=1$$.

Observe that \begin{align*} |f(z)|<1 \Longleftrightarrow |z-i|<|z-(-i)|. \end{align*} Now the last condition expresses all points $$z$$ whose distance from $$i$$ is less than its distance from $$-i$$, which are exactly the points of the upper plane.

All you need is one test point, since it maps the real axis to the unit circle. Either it maps the upper half plane to the interior or the exterior of the circle. Say take $$i$$. It goes to zero.

This is called the Cayley transformation.

For the first part, you can just check that $$1,0,-1$$ go to $$i,-1,-i$$ respectively. Then just use that Mobius transformations map generalized circles to generalized circles.

• Can you explain/give references for your first claim? Thanks – Joe Mar 29 '20 at 0:03
• Sure. These are general facts about Mobius transformations. Check Wikipedia, say. – Chris Custer Mar 29 '20 at 0:11
• You could say by connectedness, since it's continuous. – Chris Custer Mar 29 '20 at 0:16