Complex Numbers
PYP_JEE_ADV_2023_P1
Grade None
Question:
Let $z$ be a complex number satisfying $|z|^3+2z^2+4\bar{z}-8=0$, 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.
**List-I**
(P) $|z|^2$ is equal to
(Q) $|z-\bar{z}|^2$ is equal to
(R) $|z|^2+|z+\bar{z}|^2$ is equal to
(S) $|z+1|^2$ is equal to
**List-II**
(1) 12
(2) 4
(3) 8
(4) 10
(5) 7
(P)→(1) (Q)→(3) (R)→(5) (S)→(4)
(P)→(2) (Q)→(1) (R)→(3) (S)→(5)
(P)→(2) (Q)→(4) (R)→(5) (S)→(1)
(P)→(2) (Q)→(3) (R)→(5) (S)→(4)
Step-by-Step Solution
Key Concept: Separate real and imaginary parts; Im=0 forces x=1; then solve cubic in t=1+y²
Let $z=x+iy$. Imaginary part of equation:
$\text{Im}(2z^2+4\bar{z})=\text{Im}(2(x^2-y^2+2xyi)+4(x-iy))=4xy-4y=4y(x-1)=0$.
Since $y\neq0$: $x=1$.
Real part with $x=1$:
$(1+y^2)^{3/2}+2(1-y^2)+4-8=0\Rightarrow(1+y^2)^{3/2}-2y^2-2=0$.
Let $t=1+y^2$: $t^{3/2}-2t+2=0\Rightarrow t^{3/2}=2t-2\Rightarrow t(t^{1/2}-2)=0$... actually:
$t^{3/2}-2(t-1)-2=0\Rightarrow t^{3/2}-2t=0\Rightarrow t(\sqrt{t}-2)=0\Rightarrow\sqrt{t}=2\Rightarrow t=4\Rightarrow y^2=3$.
So $z=1\pm i\sqrt{3}$.
(P) $|z|^2=1+3=4$→(2).
(Q) $|z-\bar{z}|^2=|2iy|^2=4y^2=12$→(1).
(R) $|z|^2+|z+\bar{z}|^2=4+|2x|^2=4+4=8$→(3).
(S) $|z+1|^2=|2\pm i\sqrt3|^2=4+3=7$→(5).
Answer: (P)→(2),(Q)→(1),(R)→(3),(S)→(5) → B.
Correct Answer: B