# bijective and conformal function $f:H\to D$, $f(z)=\frac{z-z_0}{z-\overline{z_0}}$

$z_0 \in H:= \{z \in C: Im(z)>0\}$, $D:=\{z\in C:|z|<1\}$

Show that $f:H\to D$ , $f(z)=\frac{z-z_0}{z-\overline{z_0}}$ is bijective and conformal.

So I have to show surjective and injective and holomorphic?

Since z and $z_0$ have Im>0, $|z-z_0|<|z-\overline{z_0}|$=> $|\frac{z-z_0}{z-\overline{z_0}}|<1.$

How can I get surjective out of that?

Injective was easy. I just assumed $f(z_1)=f(z_2)$, plugged it in, and got $z_1=z_2$.

Im(z)>0 and Im($\overline{z_0}$)<0. Thus $z - \overline{z_0}\neq0$. That's why I can differentiate f? So I get

f'(z)=$\frac{z_0-\overline{z_0}}{(z-\overline{z_0})^2}$. Again $z-\overline{z_0}\neq0$. That's why it's conformal?

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Bijectivity follows from a more general theorem. A linear fractional transformation, (also known as a Moebius transformation) $$S(z)=\frac{az+b}{cz+d}$$ such that $ad-bc\neq 0$ has inverse $$S^{-1}(z)=\frac{dz-b}{-cz+a}.$$ Notice this is very similar to the theory of invertibility for $2\times 2$ matrices. In your case, $ad-bc$ amounts to $-\overline{z_0}+z_0=2i\operatorname{Im}(z_0)$. This is nonzero since $z_0$ is a point in the upper half plane.

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how can i show that $\overline{z_0}z-z_0<1$? –  Display Name Jan 23 '13 at 7:41
I have to show that $Im(f^{-1}(z))>0$, not the above, right? –  Display Name Jan 23 '13 at 7:51
@DisplayName Since $f$ is injective and maps $H$ onto $D$, it follows by definition of the inverse that $f^{-1}\colon D\to H$, in which case $\operatorname{Im}(f^{-1}(z))>0$ for $z\in D$. –  Ben West Jan 23 '13 at 8:38

The easiest way to show it is bijective is to present an inverse map: $$w=\frac{z-z_0}{z-\bar{z}_0}\Rightarrow w(z-\bar{z}_0)=z-z_0\Rightarrow z(w-1)=w\bar{z}_0-z_0\Rightarrow z=\frac{w\bar{z}_0-z_0}{w-1}$$ As $z\in \{\mathrm{Im}z>0\}$, $z-\bar{z}_0\neq0$; in the same way, as $w\in \mathbb{D}$, $w-1\neq0$. So this map $$w\mapsto\frac{w\bar{z}_0-z_0}{w-1}$$ is well defined and it is the inverse of the previous. For it being conformal, it is enough to notice that the map $f$ is holomorphic on $H$ as it is a ratio of two $1$st degree complex polynomials which do not vanish on $H$ and, as user32240 correctly pointed out, its derivative doesn't vanish: $$f'(z)=\frac{2\mathrm{Im}z_0}{(z-\bar{z}_0)^2}$$ is non-zero on the upper half-plane.

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Great answer! It helped me with a similar question I'm asking: math.stackexchange.com/questions/1245940/… –  EthanAlvaree Apr 22 at 12:50

You are right that $f$ is well-defined, but to show it is conformal you need to show it is holomorphic and its derivative is non-zero everywhere. This is not difficult since $$\frac{z-z_{0}}{z-\overline{z_{0}}}=1-\frac{2Imz_{0}}{z-\overline{z_{0}}}$$is holomorphic everywhere in the upper half plane, and its derivative is nothing but $\frac{2Imz_{0}}{(z-\overline{z_{0}})^{2}}$ which cannot vanish.

The fact that $|z-z_{0}|<|z-\overline{z_{0}}|$ follows by the geometric picture. And the picture actually showed it is surjective. If you think geometrically, obviously $f(z)$ can obtain any value in $[0,1)$ by moving along the line connecting $z_{0}$ and $\overline{z_{0}}$. Now consider the points in $\mathbb{C}$ such that $|z-z_{0}|=p|z-\overline{z_{0}}|$ with $p$ fixed, it is an eclipse. With $z$ travelling in the eclipse its angle begin to change, and when it travels back it finishes 360 degrees. So the whole region in $D$ is being covered by $f(z)$.

This is not a brutal force type proof, and if one can show this algebraically I think it would be interesting to know.

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