JEE Advance - Mathematics (2023 - Paper 1 Online - No. 17)
Let $z$ be a complex number satisfying $|z|^3+2 z^2+4 \bar{z}-8=0$, where $\bar{z}$ denotes the complex conjugate of $z$. Let the imaginary part of $z$ be nonzero.
Match each entry in List-I to the correct entries in List-II.
The correct option is:
Match each entry in List-I to the correct entries in List-II.
| 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 |
The correct option is:
$$
(P) \rightarrow(1) \quad(Q) \rightarrow(3) \quad(R) \rightarrow(5) \quad(S) \rightarrow(4)
$$
$$
(P) \rightarrow(2) \quad(Q) \rightarrow(1) \quad(R) \rightarrow(3) \quad(S) \rightarrow(5)
$$
$$
(P) \rightarrow(2) \quad(Q) \rightarrow(4) \quad(R) \rightarrow(5) \quad(S) \rightarrow(1)
$$
$$
(P) \rightarrow(2) \quad(Q) \rightarrow(3) \quad(R) \rightarrow(5) \quad(S) \rightarrow(4)
$$
Explanation
$$
\because|z|^3+2 z^2+4 \bar{z}-8=0
$$ .........(1)
Take conjugate both sides
$$ \Rightarrow|\mathrm{z}|^3+2 \overline{\mathrm{z}}^2+4 \mathrm{z}-8=0 $$ ........(2)
$$ \begin{aligned} & \text { By }(1)-(2) \\\\ & \Rightarrow 2\left(\mathrm{z}^2-\bar{z}^2\right)+4(\bar{z}-\mathrm{z})=0 \\\\ & \Rightarrow \mathrm{z}+\overline{\mathrm{z}}=2 \\\\ & \Rightarrow|\mathrm{z}+\overline{\mathrm{z}}|=2 \end{aligned} $$
Let $z=x+i y$
$$ \therefore \mathrm{x}=1 \quad \therefore \mathrm{z}=1+\mathrm{iy} $$
Put in (1)
$$ \begin{aligned} & \Rightarrow\left(1+y^2\right)^{3 / 2}+2\left(1-y^2+2 i y\right)+4(1-i y)-8=0 \\\\ & \Rightarrow\left(1+y^2\right)^{3 / 2}=2\left(1+y^2\right) \\\\ & \Rightarrow \sqrt{1+y^2}=2=|z| \end{aligned} $$
Also $\mathrm{y}= \pm \sqrt{3}$
$$ \begin{aligned} & \therefore \mathrm{z}=1 \pm \mathrm{i} \sqrt{3} \\\\ & \Rightarrow \mathrm{z}-\overline{\mathrm{z}}= \pm 2 \mathrm{i} \sqrt{3} \\\\ & \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|=2 \sqrt{3} \\\\ & \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|^2=12 \\\\ & \text { Now } \mathrm{z}+1=2+\mathrm{i} \sqrt{3} \\\\ & |\mathrm{z}+1|^2=4+3=7 \\\\ & \therefore \mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5 \\\\ & \therefore \text { Option }[\mathrm{B}] \text { is correct. } \end{aligned} $$
Take conjugate both sides
$$ \Rightarrow|\mathrm{z}|^3+2 \overline{\mathrm{z}}^2+4 \mathrm{z}-8=0 $$ ........(2)
$$ \begin{aligned} & \text { By }(1)-(2) \\\\ & \Rightarrow 2\left(\mathrm{z}^2-\bar{z}^2\right)+4(\bar{z}-\mathrm{z})=0 \\\\ & \Rightarrow \mathrm{z}+\overline{\mathrm{z}}=2 \\\\ & \Rightarrow|\mathrm{z}+\overline{\mathrm{z}}|=2 \end{aligned} $$
Let $z=x+i y$
$$ \therefore \mathrm{x}=1 \quad \therefore \mathrm{z}=1+\mathrm{iy} $$
Put in (1)
$$ \begin{aligned} & \Rightarrow\left(1+y^2\right)^{3 / 2}+2\left(1-y^2+2 i y\right)+4(1-i y)-8=0 \\\\ & \Rightarrow\left(1+y^2\right)^{3 / 2}=2\left(1+y^2\right) \\\\ & \Rightarrow \sqrt{1+y^2}=2=|z| \end{aligned} $$
Also $\mathrm{y}= \pm \sqrt{3}$
$$ \begin{aligned} & \therefore \mathrm{z}=1 \pm \mathrm{i} \sqrt{3} \\\\ & \Rightarrow \mathrm{z}-\overline{\mathrm{z}}= \pm 2 \mathrm{i} \sqrt{3} \\\\ & \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|=2 \sqrt{3} \\\\ & \Rightarrow|\mathrm{z}-\overline{\mathrm{z}}|^2=12 \\\\ & \text { Now } \mathrm{z}+1=2+\mathrm{i} \sqrt{3} \\\\ & |\mathrm{z}+1|^2=4+3=7 \\\\ & \therefore \mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 3 ; \mathrm{S} \rightarrow 5 \\\\ & \therefore \text { Option }[\mathrm{B}] \text { is correct. } \end{aligned} $$
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