JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(z = x + iy\) satisfies \(|z|-2=0\) and \(|z-i|-|z+5 i|=0\), then
- A \(x +2 y -4=0\)
- B \(x^{2}+y+4=0\)
- C \(x^{2}+2y+4=0\)
- D \(x^{2}-y+3=0\)
Answer & Solution
Correct Answer
(C) \(x^{2}+2y+4=0\)
Step-by-step Solution
Detailed explanation
\(|z-i|-|z+5 i|=0\) \(|x+(y-1) i|=|x+(y+5) i|\) \(x^{2}+(y-1)^{2}=x^{2}+(y+5)^{2}\) \((y-1)^{2}-(y+5)^{2}=0\) \((2 y+4)(-6)=0\) \(y=-2\) \(\therefore x^{2}+(-2)^{2}=4\) \(x=0\) \(Z \equiv(0,-2)\), check options
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