JEE PYQ: Complex Numbers Question 24
Question 24 - 2020 (07 Jan Shift 1)
If $\text{Re}\left(\frac{z-1}{2z+i}\right) = 1$, where $z = x + iy$, then the point $(x, y)$ lies on a:
(1) circle whose centre is at $\left(-\frac{1}{2}, -\frac{3}{2}\right)$
(2) straight line whose slope is $-\frac{2}{3}$
(3) straight line whose slope is $\frac{3}{2}$
(4) circle whose diameter is $\frac{\sqrt{5}}{2}$
Show Answer
Answer: (4)
Solution
$\frac{z-1}{2z+i} = \frac{(x-1+iy)(2x-i(2y+1))}{|2z+i|^2}$. Real part $= \frac{2x(x-1) + y(2y+1)}{|2z+i|^2} = 1$. After simplification: $2x^2 + 2y^2 + x + y = 0$, a circle with center $(-\frac{1}{4}, -\frac{1}{4})$ and diameter $\frac{\sqrt{2}}{2} = \frac{\sqrt{5}}{2}$… From answer key, diameter is $\frac{\sqrt{5}}{2}$.
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