Question

If $z_1$ and $z_2$ are two complex numbers, then the inequality $|z_1+z_2{|}^2\le (1+c)|z_1{|}^2+(1+c^{-1})|z_2{|}^2$ is true if

• A. $c\in R$
• B. $c\in (0,\infty )$
• C. $c\in R-\left\{0\right\}$
• D. $c\in (-\infty ,0)$

Answer 1

The correct option is B $c\in (0,\infty )$

$|z_1+z_2{|}^2\le (1+c)|z_1{|}^2+(1+c^{-1})|z_2{|}^2$
$[\because |z_1+z_2{|}^2=|z_1{|}^2+|z_2{|}^2+2Re(z_1\overline{z_2}\left)\right]$
$\Rightarrow |z_1{|}^2+|z_2{|}^2+2Re\left(z_1\overline{z_2}\right)\le (1+c)|z_1{|}^2+(1+c^{-1})|z_2{|}^2$
$\Rightarrow 2Re\left(z_1\overline{z_2}\right)\le c|z_1{|}^2+c^{-1}|z_2{|}^2$
$\Rightarrow c|z_1{|}^2+\frac{1}{c}|z_2{|}^2-2Re\left(z_1\overline{z_2}\right)\ge 0$
$\Rightarrow |\sqrt{c}{z}_1{|}^2+{\left|\frac{1}{\sqrt{c}}{z}_2\right|}^2-2Re\left(\sqrt{c}{z}_1\frac{1}{\sqrt{c}}\overline{z_2}\right)\ge 0$
$[\because |z_1-z_2{|}^2=|z_1{|}^2+|z_2{|}^2-2Re(z_1\overline{z_2}\left)\right]$
$\Rightarrow {|\sqrt{c}{z}_1-\frac{z_2}{\sqrt{c}}|}^2\ge 0$
which is true for $c\in (0,\infty )$