# Question about complex value functions

Given a function from the extended complex plane to itself by $$f(z)=e^{i\theta }\left(\frac{z-z_0}{z-\bar{z_0 }} \right)$$ where $$\theta$$ is real and $$z_0$$ is in the upper half plane.

I am trying to show that $$f$$ maps the upper half plane onto the open unit disc and the real line to the unit cirlce.

To do this I wrote $$f(z)=e^{i\theta } \left( \frac{|z|^2-2z_0 \text{Re}(z)+z_0 ^2 }{|z-\bar{z_0}|^2 } \right).$$

Not sure how to proceed from here.

Start with $$|f(z)|\le 1$$ for all $$z$$. Then $$|f(z)|=1$$ for real $$z$$. Finally show $$(z-z_0)(z-\bar{z_0})=re^{ia}$$ assumes all values of $$a$$ in range $$(0,2\pi)$$ for real $$z$$.

• I switched up my approach. I let $z=x+yi$ and $z_0=a+bi$ and then I got $$|f(z)|^2 =1-\frac{4by}{(x-a)^2+(y+b)^2 }$$ so I can see why $|f(z)|<1$ when $b>0$ but how to show it maps surjectively? Commented May 10, 2021 at 23:18

Here is a geometric vision.

Preliminary remark:

Let $$M_0$$ be a point of the upper half plane, $$M'_0$$ its symmetrical point wrt the real axis.

Saying that $$dist(M,M_0) ($$M$$ is closer to $$M_0$$ than to $$M'_0$$) is equivalent to say that $$M$$ belongs to the upper half plane.

(said otherwise, the real axis is the perpendicular bissector of line segment $$[M_0,M'0]$$, and point $$M$$ is on "the upper side" of it).

Let us concentrate now on

$$F(z)=\frac{z-z_0}{z-\bar{z_0 }}$$

(because the final multiplication is associated with a rotation that clearly will not impair the fact that we are in the unit disk).

We have to show that $$F(z)$$ belongs to the unit disk.

This will be done by taking the module

$$|F(z)|=\frac{|z-z_0|}{|z-\bar{z_0}|}<1$$

which amounts to say that (by associating $$M_0, M'_0$$ with $$z_0,\bar{z_0}$$, resp., with $$M'_0$$ symmetrical of $$M_0$$ wrt $$x$$ axis.)

$$\frac{dist(M,M_0)}{dist(M,M'_0)}<1 \ \iff \ dist(M,M_0)

and we are in the situation depicted in the preliminary remark.

We have thus proved the "on" part. It remains to show that this "on" is in fact an "onto", by showing that every given $$Z$$ in the (open) unit disk is reached i.e., that there exist a (unique) z such that

$$Z=\frac{z-z_0}{z-\bar{z_0 }}$$

This indeed gives rise to a first degree equation in variable $$z$$, with solution:

$$z=\dfrac{Z \overline{z_0}-z_0}{Z-1}$$

Remark: The frontier cases (real axis mapped onto unit circle) are obtained by replacing all $$<$$ sings by $$=$$ signs in the previous discussion.