JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1,\) where \(z=x+i y,\) then the point \((\mathrm{x}, \mathrm{y})\) lies on a
- A circle whose centre is at \(\left(-\frac{1}{2},-\frac{3}{2}\right)\)
- B circle whose diameter is \(\frac{\sqrt{5}}{2}\)
- C straight line whose slope is \(\frac{3}{2}\)
- D straight line whose slope is \(-\frac{2}{3}\)
Answer & Solution
Correct Answer
(B) circle whose diameter is \(\frac{\sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
\(\operatorname{Re}\left(\frac{z-1}{2 z+i}\right)=1\) Put \(z=x+i y\) \(\operatorname{Re}\left(\frac{(\mathrm{x}+\mathrm{iy})-1}{2(\mathrm{x}+\mathrm{i} \mathrm{y})+\mathrm{i}}\right)=1\)…
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