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JEE Mains · Maths · STD 11 - 4.1 complex nubers
The equation \( lm\,\left( {\frac{{iz - 2}}{{z - i}}} \right) + 1 = 0\,,z \in C\,,z \ne i\) represents a part of a circle having radius equal to
- A \(2\)
- B \(1\)
- C \(\frac{3}{4}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
Let \(z=x+y\) \(\operatorname{Im}\left[\left(\frac{i x-y-2}{x+(y-1\,i)}\right)\left(\frac{x-(y-1\,i)}{x-(y-1\,i)}\right)\right]+1=0\) On solving, we get: \({ = 2{x^2} + 2{y^2} - y - 1 = 0}\) \({ = {x^2} + {y^2} - 1/2y - 1/2 = 0}\)…
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