If $z=x+i y$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,…
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
- $x$-axis
- $y$-axis
- $y=x$
- $2y=x$
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
Asked in: AP EAMCET 2002
Practice more Complex Number questions on Aicharya