Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$Z_1=a+bi$

$Z_2=c+di$

By given condition: $a^2+b^2<1<c^2+d^2$

To prove: $|1-Z_1{Z}_2|<|Z_1-Z_2|$

$\mid 1-(a+bi)(c-di)\mid <\mid a-c+(b-d)i\mid$

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

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

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

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

$bd<abcd$ Since its all positive

Hence LHS = RHS

so this is TRUE