JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(z\) be a complex number such that \(\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i\). Then \(z\) lies on the circle of radius \(2\) and centre
- A \((2,0)\)
- B \((0,0)\)
- C \((0,2)\)
- D \((0,-2)\)
Answer & Solution
Correct Answer
(D) \((0,-2)\)
Step-by-step Solution
Detailed explanation
\(( z -2 i )(\overline{ z }+2 i )=4( z + i )(\overline{ z }- i )\) \(z \overline{ z }+4+2 i ( z -\overline{ z })=4( z \overline{ z }+1+ i (\overline{ z }- z ))\) \(3 z \overline{ z }-6 i ( z -\overline{ z })=0\) \(x ^2+ y ^2-2 i (2 iy )=0\) \(x ^2+ y ^2+4 y =0\)
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