Question:
If \( {\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1 \), where z = x + iy, then the point (x, y) lies on a :
If \( {\mathop{\rm Re}\nolimits} \left( {{{z - 1} \over {2z + i}}} \right) = 1 \), where z = x + iy, then the point (x, y) lies on a :
Show Hint
Multiply numerator and denominator by the conjugate of \(2z+i\), take the real part and equate to 1; then complete the square.
Updated On: Oct 31, 2026
- straight line whose slope is \( {3 \over 2} \)
- straight line whose slope is \( -{2 \over 3} \)
- circle whose diameter is \( {{\sqrt 5 } \over 2} \)
- circle whose centre is at \( \left( { - {1 \over 2}, - {3 \over 2}} \right) \)
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The Correct Option is C
Solution and Explanation
Let z = x + iy. Then \(\dfrac{z-1}{2z+i}=\dfrac{(z-1)(2\bar z-i)}{|2z+i|^2}\).
Real part of numerator: \((x-1)(2x)+y(2y+1)=2x^2+2y^2-2x+y\). Denominator: \(|2z+i|^2=4x^2+(2y+1)^2=4x^2+4y^2+4y+1\).
Setting the ratio equal to 1: \(2x^2+2y^2-2x+y=4x^2+4y^2+4y+1\Rightarrow 2x^2+2y^2+2x+3y+1=0\).
So \(x^2+y^2+x+\tfrac32y+\tfrac12=0\), a circle with centre \(\left(-\tfrac12,-\tfrac34\right)\) and \(r^2=\tfrac14+\tfrac9{16}-\tfrac12=\tfrac5{16}\), so \(r=\tfrac{\sqrt5}{4}\) and diameter \(\tfrac{\sqrt5}{2}\).
Answer: (C) circle whose diameter is \(\frac{\sqrt5}{2}\)
Real part of numerator: \((x-1)(2x)+y(2y+1)=2x^2+2y^2-2x+y\). Denominator: \(|2z+i|^2=4x^2+(2y+1)^2=4x^2+4y^2+4y+1\).
Setting the ratio equal to 1: \(2x^2+2y^2-2x+y=4x^2+4y^2+4y+1\Rightarrow 2x^2+2y^2+2x+3y+1=0\).
So \(x^2+y^2+x+\tfrac32y+\tfrac12=0\), a circle with centre \(\left(-\tfrac12,-\tfrac34\right)\) and \(r^2=\tfrac14+\tfrac9{16}-\tfrac12=\tfrac5{16}\), so \(r=\tfrac{\sqrt5}{4}\) and diameter \(\tfrac{\sqrt5}{2}\).
Answer: (C) circle whose diameter is \(\frac{\sqrt5}{2}\)
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