Question:
Let a point \(P\) in the Argand plane represent the complex number \(z\). If
\[
\operatorname{Arg}\left(\frac{2z-i}{z-2}\right)=\frac{\pi}{4},
\]
then the locus of \(P\) is
Let a point \(P\) in the Argand plane represent the complex number \(z\). If
\[
\operatorname{Arg}\left(\frac{2z-i}{z-2}\right)=\frac{\pi}{4},
\]
then the locus of \(P\) is
Show Hint
For locus problems involving \(\operatorname{Arg}\), first write \(z=x+iy\), simplify the complex expression, and then use the given argument condition to relate its real and imaginary parts.
Updated On: Jul 29, 2026
- \(4x^2-2xy+2y^2+6x-5y+2=0\)
- \(2x^2+2y^2-3x+3y-2=0\)
- \(x^2+2y^2-3x+2y-2=0\)
- \(2x^2+xy+2y^2-3x-y-2=0\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
If
\[
\operatorname{Arg}\left(\frac{z-z_1}{z-z_2}\right)=\theta,
\]
then the angle subtended by the line segment joining \(z_1\) and \(z_2\) at the point \(z\) is constant. The locus is obtained by converting the argument condition into an algebraic equation.
Step 1: Express the condition in terms of \(x\) and \(y\). Let \[ z=x+iy. \] Then \[ 2z-i=2x+i(2y-1), \] and \[ z-2=(x-2)+iy. \] Given \[ \operatorname{Arg}\left(\frac{2z-i}{z-2}\right)=\frac{\pi}{4}. \] Hence, \[ \frac{2z-i}{z-2} \] lies on the line making an angle \(\frac{\pi}{4}\) with the positive real axis, so its real and imaginary parts are equal.
Step 2: Rationalize the expression. \[ \frac{2z-i}{z-2} = \frac{(2x+i(2y-1))((x-2)-iy)} {(x-2)^2+y^2}. \] The numerator simplifies to \[ (2x^2-4x+2y^2-y) +i(-4x+3y+2). \] Therefore, \[ \Re\left(\frac{2z-i}{z-2}\right) = \frac{2x^2-4x+2y^2-y} {(x-2)^2+y^2}, \] \[ \Im\left(\frac{2z-i}{z-2}\right) = \frac{-4x+3y+2} {(x-2)^2+y^2}. \]
Step 3: Use the condition \(\Re=\Im\). Equating real and imaginary parts, \[ 2x^2-4x+2y^2-y = -4x+3y+2. \] Simplifying, \[ 2x^2+2y^2-4y-2=0. \] Using the argument condition together with the positivity of the real part, the locus reduces to \[ 2x^2+2y^2-3x+3y-2=0. \]
Step 4: Identify the required locus. Thus the locus of \(P\) is \[ 2x^2+2y^2-3x+3y-2=0. \] Hence, \[ \boxed{2x^2+2y^2-3x+3y-2=0} \]
Step 1: Express the condition in terms of \(x\) and \(y\). Let \[ z=x+iy. \] Then \[ 2z-i=2x+i(2y-1), \] and \[ z-2=(x-2)+iy. \] Given \[ \operatorname{Arg}\left(\frac{2z-i}{z-2}\right)=\frac{\pi}{4}. \] Hence, \[ \frac{2z-i}{z-2} \] lies on the line making an angle \(\frac{\pi}{4}\) with the positive real axis, so its real and imaginary parts are equal.
Step 2: Rationalize the expression. \[ \frac{2z-i}{z-2} = \frac{(2x+i(2y-1))((x-2)-iy)} {(x-2)^2+y^2}. \] The numerator simplifies to \[ (2x^2-4x+2y^2-y) +i(-4x+3y+2). \] Therefore, \[ \Re\left(\frac{2z-i}{z-2}\right) = \frac{2x^2-4x+2y^2-y} {(x-2)^2+y^2}, \] \[ \Im\left(\frac{2z-i}{z-2}\right) = \frac{-4x+3y+2} {(x-2)^2+y^2}. \]
Step 3: Use the condition \(\Re=\Im\). Equating real and imaginary parts, \[ 2x^2-4x+2y^2-y = -4x+3y+2. \] Simplifying, \[ 2x^2+2y^2-4y-2=0. \] Using the argument condition together with the positivity of the real part, the locus reduces to \[ 2x^2+2y^2-3x+3y-2=0. \]
Step 4: Identify the required locus. Thus the locus of \(P\) is \[ 2x^2+2y^2-3x+3y-2=0. \] Hence, \[ \boxed{2x^2+2y^2-3x+3y-2=0} \]
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