Question:medium

If \[ \arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}, \] then the locus of the point \(P(z)\) on the Argand plane is a

Show Hint

Remember the standard result: \[ \arg\left(\frac{z-z_1}{z-z_2}\right)=\alpha \] represents the locus of points from which the segment joining \(z_1\) and \(z_2\) is seen under a constant angle \(\alpha\). Therefore, the locus is a circle.
Updated On: Jun 16, 2026
  • line
  • circle
  • parabola
  • hyperbola
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Read the geometry of the argument.
The expression $\arg\left(\dfrac{z-1}{z+1}\right)$ is the angle at the point $z$ between the lines drawn from $z$ to the fixed points $1$ and $-1$. Here that angle is fixed at $\tfrac{\pi}{4}$.

Step 2: Use the constant-angle theorem.
A point that sees a fixed segment under a constant angle traces an arc of a circle (the inscribed-angle idea). The segment here joins $z_1=1$ and $z_2=-1$.

Step 3: Confirm with coordinates.
Write $z=x+iy$. Then $\dfrac{z-1}{z+1}=\dfrac{(x-1)+iy}{(x+1)+iy}$. Its argument being $\tfrac{\pi}{4}$ means the tangent of the angle equals $1$.

Step 4: Form the tangent condition.
Computing the argument gives $\dfrac{\text{Imaginary part}}{\text{Real part}}=1$, which leads to an equation containing $x^2+y^2$ terms with equal coefficients.

Step 5: Recognise the curve.
Because $x^2$ and $y^2$ appear with the same coefficient and there is no $xy$ term, the equation is that of a circle, not a line, parabola or hyperbola.

Step 6: State the locus.
So the point $P(z)$ moves on a circle passing through $1$ and $-1$. \[ \boxed{\text{circle}} \]
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