JEE Mains · Maths · STD 11 - 4.1 complex nubers
For all complex numbers \(z\) of the form \(1 + i\alpha\), \(\alpha \in R\) , if \(z^2\, = x + iy\), then
- A \(y^2 -4x +2\,=0\)
- B \(y^2 +4x -4\,=0\)
- C \(y^2-4x -4\,=0\)
- D \(y^2 +4x +2\,=0\)
Answer & Solution
Correct Answer
(B) \(y^2 +4x -4\,=0\)
Step-by-step Solution
Detailed explanation
Let \(z=1+i \alpha, \alpha \in R\) \(z^{2}=(1+i \alpha)(1+i \alpha)\) \(x+i y=\left(1+2 i \alpha-\alpha^{2}\right)\) On comparing real and imaginary parts, we get \(x=1-\alpha^{2}, y=2 \alpha\) Now, consider option \((b)\), which is \(y^{2}+4 x-4=0\) \(LHS\) : \(y^{2}+4 x-4\)…
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