Question

If $z_1$ and $z_2$ are two complex numbers such that $|z_1|<1<|z_2|$ then prove that $\left|\frac{1-z_1\overline{z_2}}{z_1-z_2}\right|<1$.

Answer 1

$z_1=a+bi$

$z_2=c+di$

By given condition:

$a^2+b^2<1<c^2+d^2$

To prove: $|1-z_1\overline{z_2}|<|z_1-z_2|$

$:|1-(a+bi)(c-di)|<|a-c+(b-d)i|$

$:|1-ac+bd+(ad-bc)i|<\sqrt{(a-c{)}^2+(b-d{)}^2}$

$:(1-ac+bd{)}^2+(ad-bc{)}^2<(a-c{)}^2+(b-d{)}^2$

$:1+(a^2+b^2)(c^2+d^2)+4bd-4abcd<a^2+b^2+c^2+d^2$

$(a^2+b^2)(c^2+d^2)<a^2+b^2$ From given first half of inequality

$1<c^2+d^2$ From second half

$bd<abcd$ Since its all positive

$⟹LHS<c^2+d^2+(a^2+b^2)+0=RHS$

Hence proved