Let z be complex number satisfying |z|^3+2 z^ | vedclass

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Question

$\because|z|^3+2 z^2+4 \bar{z}-8=0$     $. . . . . . (1)$

Take conjugate both sides

$\Rightarrow|z|^3+2 z^2+4 z-8=0$

Vedclass · Accepted Answer

$(\mathrm{A})(\mathrm{P}) \rightarrow(1)(\mathrm{Q}) \rightarrow(3)(\mathrm{R}) \rightarrow(5)(\mathrm{S}) \rightarrow(4)$

Vedclass · Answer

$\because|z|^3+2 z^2+4 \bar{z}-8=0$     $. . . . . . (1)$

Take conjugate both sides

$\Rightarrow|z|^3+2 z^2+4 z-8=0$    $. . . . . . .(2)$

$\operatorname{By}(1)-(2)$

$\Rightarrow 2\left(z^2-\bar{z}^2ight)+4(\bar{z}-z)=0$

$\Rightarrow z+\bar{z}=2$     $. . . . . .(3)$

$\Rightarrow|z+\bar{z}|=2$   $. . . . . (4)$

Let $z=x+i y$

$	herefore x=1 \quad 	herefore z=1+\mathrm{i}$

Put in $(1)$

$\Rightarrow\left(1+y^2ight)^{3 / 2}+2\left(1-y^2+2 i yight)+4(1-i y)-8=0$

$\Rightarrow\left(1+y^2ight)^{3 / 2}=2\left(1+y^2ight)$

$\Rightarrow \sqrt{1+y^2}=2=|z|$

Also $\mathrm{y}= \pm \sqrt{3}$

$	herefore z=1 \pm i \sqrt{3}$

$\Rightarrow z-\bar{z}= \pm 2 i \sqrt{3}$

$\Rightarrow|z-\bar{z}|=2 \sqrt{3}$

$\Rightarrow|z-\bar{z}|^2=12$

Now $z+1=2+i \sqrt{3}$

$|z+1|^2=4+3=7$

$	herefore P ightarrow 2 ; Q ightarrow 1 ; R ightarrow 3 ; S ightarrow 5$

$	herefore$ Option $[B]$ is correct.

List-$I$	List-$II$
($P$) $\|z\|^2$ is equal to	($1$) $12$
($Q$) $\|z-\bar{z}\|^2$ is equal to	($2$) $4$
($R$) $\|z\|^2+\|z+\bar{z}\|^2$ is equal to	($3$) $8$
($S$) $\|z+1\|^2$ is equal to	($4$) $10$
	($5$) $7$

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