If $\ z_1$ and $\ z_2$ are two complex numbers such that | $\ z_1$ | < 1 < | $\ z_2$ | then prove that |$(\frac{1-\ z_1 \overline z_2}{\ z_1 -\ z_2})$ |<1
My initial approach $\ z_1=\ r_1 e^{i\alpha}$ where $\ r_1$ $\in$ (0,1) and $\ z_2=\ r_2 e^{i\beta}$ where $\ r_2$ >1
After this step i tried to evaluate it but not able to proceed
Apologies in advance for lack of a figure, but you can show this geometrically. Denote $z_1, z_2, z_1\overline{z_2}, 1$ by $A,B,C,D$ respectively.
Consider $\triangle OAB, \triangle OCD$. where $O$ is the origin. We observe the following:
$\angle AOB = |\arg z_1 - \arg z_2| = \angle COD $
$OD = 1 <|z_2|$ (given)
$OC =|z_1\overline{z_2}| = |z_1||z_2|<|z_1| \ \ (\because |z_1|<1)$
This means $\triangle OAB, \triangle OCD$ share a vertex angle, and the two sides of $\triangle OCD$ that include this vertex angle are respectively less than the sides of $\triangle OAB$.
That implies $CD<AB$ or $|1-z_1\overline{z_2}|<|z_1-z_2|$
Hint: Use $|z|^2=z\bar{z}$. Try to prove that $|1-z_1\bar{z}_2|^2<|z_1-z_2|^2$.
