Question

If $z_1$ and $z_2$ are two complex numbers and $c>0$, then $|z_1-z_2{|}^2\ge (1-c)|z_1{|}^2+(1-\frac{1}{c})|z_2{|}^2$ is

• A. True for all $z_1,z_2\in C$
• B. True is $z_1=\overline{z_2}$
• C. False for all $z_1,z_2\in C$
• D. True if $z_1/z_2$ is real

Answer 1

Answer: (A)

True for all $z_1,z_2\in C$

Consider, $c|z_1{|}^2+\frac{|z_2{|}^2}{c}\ge 2\sqrt{|z_1{|}^2|z_2{|}^2}$ .....[ A.M $\ge$ G.M]
$\Rightarrow c|z_1{|}^2+\frac{|z_2{|}^2}{c}\ge 2|z_1||z_2|$
$\Rightarrow -2|z_1||z_2|\ge -c|z_1{|}^2-\frac{|z_2{|}^2}{c}$
$\Rightarrow |z_1{|}^2+|z_2{|}^2-2|z_1||z_2|\ge |z_1{|}^2+|z_2{|}^2-c|z_1{|}^2-\frac{|z_2{|}^2}{c}$
$\Rightarrow {(|z_1|-|z_2|)}^2\ge (1-c)|z_1{|}^2+(1-\frac{1}{c})|z_2{|}^2$
Since, $|z_1-z_2|\ge |z_1|-|z_2|$
Therefore, ${|z_1-z_2|}^2\ge (1-c)|z_1{|}^2+(1-\frac{1}{c})|z_2{|}^2$
Therefore, above inequality is true for all $z_1$ and $z_2$.
Ans: A