Question

The locus of a point $ z $ satisfying: $$ |z|^2 = \text{Re}(z) $$ is a circle with centre:

• A. \( \left(0, \frac{1}{2} \right) \)
• B. \( \left(-\frac{1}{2}, 0 \right) \)
• C. \( \left(\frac{1}{2}, 0 \right) \)
• D. \( \left(0, -\frac{1}{2} \right) \)

Hint:

To convert a complex equation to a geometric locus, write \( z = x + iy \), then apply algebra and complete the square if needed.

Answer 1

The Correct Option is C

Let \( z = x + iy \). Then: \[ |z|^2 = x^2 + y^2,\quad \text{Re}(z) = x \] So the equation becomes: \[ x^2 + y^2 = x \Rightarrow x^2 - x + y^2 = 0 \] Complete the square: \[ x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4} \Rightarrow \left(x - \frac{1}{2}\right)^2 + y^2 = \left(\frac{1}{2}\right)^2 \] This is the equation of a circle with:
- Centre \( \left(\frac{1}{2}, 0 \right) \)
- Radius \( \frac{1}{2} \)