JEE Advanced · Mathematics · 3. Complex Numbers
Let be a complex number satisfying , where denotes the complex conjugate of . Let the imaginary part of be nonzero.
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
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
Step-by-step Solution
Detailed explanation
Given,
Now on taking conjugate both side we get,
Now subtracting both equations we get,
(not possible) or
So,
Now putting the value of 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) \(|z+1|^2=|1+\lambda i+1|^2=4~+\) \(\lambda^2=4+3=7\)
\(\therefore P \rightarrow(2), Q \rightarrow(1), R \rightarrow(3), S \rightarrow(5)\)
Now on taking conjugate both side we get,
Now subtracting both equations we get,
(not possible) or
So,
Now putting the value of 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) \(|z+1|^2=|1+\lambda i+1|^2=4~+\) \(\lambda^2=4+3=7\)
\(\therefore P \rightarrow(2), Q \rightarrow(1), R \rightarrow(3), S \rightarrow(5)\)
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