Question

If $z_1$ and $z_2$ are two complex numbers such that $\left|z_1\right|<1<\left|z_2\right|$, then show that $\left|\frac{(1-z_1{\overline{z}}_2)}{(z_1-z_2)}\right|<1$.

Answer 1

Given $\left|z_1\right|<1$ and $\left|z_2\right|>1$ (1)

Then, to prove

$\left|\frac{1-z_1\overline{z_2}}{z_1-z_2}\right|<1$ $(\because \left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|})$

$\Rightarrow |1-z_2\overline{z_2}|<|z_1-z_2|$ ...(2)

On squaring both, sides we get

$(1-z_1\overline{z_2})(1-\overline{z_1}{z}_2)<(z_1-z_2)(\overline{z_1}-\overline{z_2})(\because {\left|z\right|}^2=z\overline{z})\Rightarrow 1+{\left|z_1\right|}^2{\left|z_2\right|}^2<{\left|z_1\right|}^2+{\left|z_2\right|}^2\Rightarrow 1-{\left|z_1\right|}^2-{\left|z_2\right|}^2{+\left|z_2\right|}^2{\left|z_2\right|}^2<0$

$\Rightarrow (1-{\left|z_1\right|}^2)(1-{\left|z_2\right|}^2)<0$ ...(3)

Which is true by equation (1)

$\left|z_1\right|<1$ and $\left|z_2\right|>1$

$\therefore (1-{\left|z_1\right|}^2)>0$ and $(1-{\left|z_2\right|}^2)<0$

Therefore equation (3) is true whenever equation (2) is true

$\Rightarrow \left|\frac{1-z_2\overline{z_2}}{z_1-z_2}\right|<1$