Q137 / 200 · AP EAMCET 2002 · Mathematics

If $z=x+i y$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$, then the locus of $z$ is

  • Correct answer
    A
    $x$-axis
  • B
    $y$-axis
  • C
    $y=x$
  • D
    $2y=x$

Solution

We have, $$ \left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2 $$ $$ \begin{array}{rlrl} \Rightarrow & \left|x+i y+\frac{i}{2}\right|^2 & =\left|x+i y-\frac{i}{2}\right|^2 \\ \Rightarrow & \left|x+i\left(y+\frac{1}{2}\right)\right|^2 & =\left|x+i\left(y+\frac{1}{2}\right)\right|^2 \\ \Rightarrow & x^2+\left(y+\frac{1}{2}\right)^2 & =x^2+\left(y-\frac{1}{2}\right)^2 \\ \Rightarrow & x^2+y^2+\frac{1}{4}+y & =x^2+y^2+\frac{1}{4}-y \\ \Rightarrow & & 2 y & =0 \\ \Rightarrow & y & =0 \end{array} $$ Locus of $z$ is $x$-axis.

Hence, the correct answer is option: 'A'.

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