Question

If $z_1,z_2$ are two complex numbers and $c>0$ such that ${|z_1+z_2|}^2\le (1+c){\left|z_1\right|}^2+k{\left|z_2\right|}^2,$ then $k=$

• A. $1-c$
• B. $c-1$
• C. $1+c^{-1}$
• D. $1-c^{-1}$

Answer 1

The correct option is C $1+c^{-1}$

Since Re $\left(z_1{\overline{z}}_2\right)\le \left|z_1{\overline{z}}_2\right|$

$\therefore {\left|z_1\right|}^2+{\left|z_2\right|}^2+Re\left(z_1{\overline{z}}_2\right)\le {\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1{\overline{z}}_2\right|$

$\Rightarrow {|z_1+z_2|}^2\le {\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|$ ...(1)

Also, Since $A.M.\ge G.M.$

$\therefore \frac{{\left(\sqrt{c}\left|z_1\right|\right)}^2+{\left(\frac{1}{\sqrt{c}}\left|z_2\right|\right)}^2}{2}\ge {\{c.{\left|z_1\right|}^2.\frac{1}{c}{\left|z_2\right|}^2\}}^{\frac{1}{2}}(\because c>0)$

$\Rightarrow c{\left|z_1\right|}^2+\frac{1}{c}{\left|z_1\right|}^2\ge 2\left|z_1\right|\left|z_2\right|$

$\therefore {\left|z_1\right|}^2+{\left|z_1\right|}^2+2\left|z_1\right|\left|z_2\right|\le {\left|z_1\right|}^2+{\left|z_2\right|}^2+c{\left|z_1\right|}^2+\frac{1}{c}{\left|z_2\right|}^2$

$\Rightarrow {\left|z_1\right|}^2+{\left|z_2\right|}^2+2\left|z_1\right|\left|z_2\right|\le (1+c^{-1})\left({\left|z_2\right|}^2\right)$ ...(2)

From (1) and (2), we get

${|z_1+z_2|}^2\le (1+c){\left|z_1\right|}^2+(1+c^{-1}){\left|z_2\right|}^2$

$\therefore k=1+c^{-1}$