The image of a Möbius transformation

Question

I have the following problem it seems easy but I couldn't find the answer!

For the following $m(z)$ $=$ $\frac {z - i}{iz-1 }$ show that $m$($\mathbb{H}$) $=$ $\mathbb{D}$

I tried to show that |$m(z)$| $\leq$ $1$ for $z$ $\in$ $\mathbb{H}$ and I got the following after multiply with the conjugate of denominator

-$i$ |$z^2$| $-$ $2$$Re(z)$ $+$ $i$ $/$ |$z^2$| $+$ $2$ $Im(z)$ $+$ $1$

But I don't know how I can show that this value is less than or equal to $1$ any help please

Answer 1

$$ |m(z)|\leq 1 \iff |z-i|\leq |iz-1| $$ Note that the RHS can be rewritten as $$ |iz-1|= |-i| |iz-1| = |(-i)(iz-1)| = |z+1| $$ And then you are comparing $|z-i|\overset{?}{\leq} |z+i|$. Expand $z=x+iy$ in real and imaginary part and instead of proving the above inequality, prove the inequality for squares (they are equivalent) $$ |z-i|^2 \overset{?}{\leq} |z+i|^2 $$ and you can compute both sides: $$ LHS=|z-i|^2 = x^2+(y-1)^2 $$ $$ RHS=|z+i|^2 = x^2+(y+1)^2 $$ Note that $LHS\leq RHS \iff x^2+y^2+1-2y \leq x^2 +y^2+1 +2y$. Since $y\geq 0$ since $z\in\mathbb{H}$ you are done.

P.S: this $m$ is called the Cayley transform between the Poincaré half plane and disc models of hyperbolic geometry.

Answer 2

Mobius transformations map extended lines and circles to extended lines and circles. So $m(\partial H)$ is either an extended line or a circle. To find out which it is, plug in three points: $m(0) = i$, $m(1) = \frac{1 - i}{i - 1} = -1$, $m(-1) = 1$. Hence $m(\partial H) = \partial{D}$. Now a connectedness argument shows that either $m(H) = D$ or $m(H) = \overline{\mathbb{C}} \setminus {\overline{D}}$. To find out which it is, plug in a point: $m(i) = 0 \in D$. Hence $m(H) = D$.