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AP EAMCET · Maths · Complex Number

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

  1. A \(\left(0, \frac{1}{2}\right)\)
  2. B \(\left(-\frac{1}{2}, 0\right)\)
  3. C \(\left(\frac{1}{2}, 0\right)\)
  4. D \(\left(0,-\frac{1}{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left(\frac{1}{2}, 0\right)\)

Step-by-step Solution

Detailed explanation

Let \(z=x+i y\) \(|z|=\sqrt{x^2+y^2}\) Now, \(|z|^2=\operatorname{Re}(z)\) \(x^2+y^2=x\) \(\Rightarrow \quad x^2+y^2-x=0\) \(g=1 / 2, f=0\) So, centre of circle \(\left(\frac{1}{2}, 0\right)\).