If z 1 and z 2 are two complex numbers- then the inequality - z 1- z 2-2-1- c - z 1-2-1- c -1- z 2-2 is true ifA- c - RB- c -0- -C- c - R -0D- c - 0

|z_1+z_2|^2 ≤ (1+c) |z_1|^2 +(1+ c^ 1) |z_2|^2 [∵|z_1+z_2|^2=|z_1|^2+|z_2|^2+2Re(z_1z_2)] ⇒ |z_1|^2+ |z_2|^2+ 2Re(z_1z_2) ≤ (1+c) |z_1|^2+ (1+ c^ 1) |z_2|^2 ⇒ 2 ...

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Standard XII

Mathematics

Inequality of 2 Complex Numbers

If z 1 and z ...

Question

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

c \in R

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c \in (0, \infty)

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c \in R - {0}

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c \in (- \infty, 0)

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Solution

The correct option is B $c \in (0, \infty)$
$| z_{1} + z_{2} |^{2} \leq (1 + c) | z_{1} |^{2} + (1 + c^{- 1}) | z_{2} |^{2}$
$[∵ | z_{1} + z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} + 2 R e (z_{1}_{2})]$
$\Rightarrow | z_{1} |^{2} + | z_{2} |^{2} + 2 R e (z_{1}_{2}) \leq (1 + c) | z_{1} |^{2} + (1 + c^{- 1}) | z_{2} |^{2}$
$\Rightarrow 2 R e (z_{1}_{2}) \leq c | z_{1} |^{2} + c^{- 1} | z_{2} |^{2}$
$\Rightarrow c | z_{1} |^{2} + \frac{1}{c} | z_{2} |^{2} - 2 R e (z_{1}_{2}) \geq 0$
$\Rightarrow | \sqrt{c} z_{1} |^{2} + {∣ ∣ ∣ \frac{1}{\sqrt{c}} z_{2} ∣ ∣ ∣}_{2}^{2} - 2 R e (\sqrt{c} z_{1} \frac{1}{\sqrt{c}}_{2}) \geq 0$
$[∵ | z_{1} - z_{2} |^{2} = | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (z_{1}_{2})]$
$\Rightarrow {∣ ∣ ∣ \sqrt{c} z_{1} - \frac{z_{2}}{\sqrt{c}} ∣ ∣ ∣}_{1}^{2} \geq 0$
which is true for $c \in (0, \infty)$

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If z1 and z2 are two complex numbers, then the inequality |z1+z2|2≤(1+c)|z1|2+(1+c−1)|z2|2 is true if

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