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JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(a = lm\left( {\frac{{1 + {z^2}}}{{2iz}}} \right)\), where \(z\) is any non-zero complex number. The set \(A = \{ a:\left| z \right| = 1\,and\,z \ne \pm 1\} \) is equal to
- A \(\left( { - 1,1} \right)\)
- B \(\left[ { - 1,1} \right]\)
- C \(\left[ {0,1} \right)\)
- D \(\left( { - 1,0} \right]\)
Answer & Solution
Correct Answer
(A) \(\left( { - 1,1} \right)\)
Step-by-step Solution
Detailed explanation
Let \(z=x+i y \Rightarrow z^{2}=x^{2}-y^{2}+2 i x y\) Now, \(\frac{{1 + {z^2}}}{{2iz}} = \) \({ = \frac{{1 + {x^2} - {y^2} + 2ixy}}{{2i(x + iy)}}}\) \({ = \frac{{\left( {{x^2} - {y^2} + 1} \right) + 2ixy}}{{2ix - 2y}}}\)…
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