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

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(B) Given equation: $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$  ...$(1)$Taking the conjugate of both sides:$|z|^3 + 2\bar{z}^2 + 4z - 8 = 0$  ...$(2)$Subtracti

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$(B) (P) \rightarrow (2), (Q) \rightarrow (1), (R) \rightarrow (3), (S) \rightarrow (5)$

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(B) Given equation: $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$  ...$(1)$Taking the conjugate of both sides:$|z|^3 + 2\bar{z}^2 + 4z - 8 = 0$  ...$(2)$Subtracting $(2)$ from $(1)$:$2(z^2 - \bar{z}^2) + 4(\bar{z} - z) = 0$$2(z - \bar{z})(z + \bar{z}) - 4(z - \bar{z}) = 0$Since $	ext{Im}(z) 
eq 0$,$z - \bar{z} 
eq 0$,so $2(z + \bar{z}) - 4 = 0 \Rightarrow z + \bar{z} = 2$.Let $z = x + iy$. Then $2x = 2 \Rightarrow x = 1$.Substituting $x=1$ into the original equation $|z|^3 + 2z^2 + 4\bar{z} - 8 = 0$:$|z|^3 + 2(1+iy)^2 + 4(1-iy) - 8 = 0$$|z|^3 + 2(1 - y^2 + 2iy) + 4 - 4iy - 8 = 0$$|z|^3 + 2 - 2y^2 + 4iy + 4 - 4iy - 8 = 0$$|z|^3 - 2y^2 - 2 = 0$Since $|z|^2 = x^2 + y^2 = 1 + y^2$,we have $|z|^3 = (1+y^2)^{3/2}$.$(1+y^2)^{3/2} - 2(y^2 + 1) = 0$$(1+y^2) [\sqrt{1+y^2} - 2] = 0$Since $1+y^2 
eq 0$,$\sqrt{1+y^2} = 2 \Rightarrow |z| = 2$.Thus,$|z|^2 = 4$.$1 + y^2 = 4 \Rightarrow y^2 = 3 \Rightarrow y = \pm \sqrt{3}$.$|z-\bar{z}|^2 = |2iy|^2 = 4y^2 = 4(3) = 12$.$|z|^2 + |z+\bar{z}|^2 = 4 + |2|^2 = 4 + 4 = 8$.$|z+1|^2 = |(1+iy)+1|^2 = |2+iy|^2 = 2^2 + y^2 = 4 + 3 = 7$.Therefore,$P ightarrow 2, Q ightarrow 1, R ightarrow 3, S ightarrow 5$.

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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