ExamBro
ExamBro
JEE Advanced · Mathematics · 3. Complex Numbers

Let z be a complex number satisfying z3+2z2+4z¯-8=0, where 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 - IList - 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

  1. A P1 Q3 R5 S4
  2. B P2 Q1 R3 S5
  3. C P2 Q4 R5 S1
  4. D P2 Q3 R5 S4
Verified Solution

Answer & Solution

Correct Answer

(B) P2 Q1 R3 S5

Step-by-step Solution

Detailed explanation

Given,
z3+2z2+4z¯-8=0 .......1
Now on taking conjugate both side we get,
z3+2z¯2+4z-8=0 ........2
Note z¯=z
Now subtracting both equations we get,
2z2-z¯2+4z¯-z=0
z-z¯2z+z¯-4=0
 z=z¯ (not possible) or 4x=4x=1 as z+z¯=2x
So, z=1+λi
z=1+λ2 & z¯=1-λi
Now putting the value of z, z & z¯ in given equation we get,
\(\left(1+\lambda^2\right)^{3 / 2}+2\left(1-\lambda^2+2 \lambda i\right)\) \(+~4(1-\lambda i)-8=0 \)
\( \Rightarrow\left(1+\lambda^2\right)^{3 / 2}+2\left(1-\lambda^2\right)=4 \)
\( \Rightarrow\left(1+\lambda^2\right)^{3 / 2}=2\left(1+\lambda^2\right) \)
\( \Rightarrow\left(1+\lambda^2\right)\left[\sqrt{1+\lambda^2}-2\right]=0 \)
\( \Rightarrow \lambda^2=3\)
Now solving,
(P) \(|z|^2=1+\lambda^2=1+3=4\)
(Q) \(|z-\bar{z}|^2=|1+\lambda i-(1-\lambda i)|^2\) \(=|2 \lambda i|^2=4 \lambda^2=12\)
(R) \(|z|^2+|z+\bar{z}|^2=4+|(1+\lambda i)\) \(+~(1-\lambda i)|^2=4+4=8\)
(S) \(|z+1|^2=|1+\lambda i+1|^2=4~+\) \(\lambda^2=4+3=7\)
\(\therefore P \rightarrow(2), Q \rightarrow(1), R \rightarrow(3), S \rightarrow(5)\)
Same subject
Explore more questions on app
From JEE Advanced
Explore more questions on app