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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
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Verified Answer
The correct answer is:
$x$-axis
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.
$$
\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2
$$
$$
\begin{array}{rlrl}
\Rightarrow & \left|x+i y+\frac{i}{2}\right|^2 & =\left|x+i y-\frac{i}{2}\right|^2 \\
\Rightarrow & \left|x+i\left(y+\frac{1}{2}\right)\right|^2 & =\left|x+i\left(y+\frac{1}{2}\right)\right|^2 \\
\Rightarrow & x^2+\left(y+\frac{1}{2}\right)^2 & =x^2+\left(y-\frac{1}{2}\right)^2 \\
\Rightarrow & x^2+y^2+\frac{1}{4}+y & =x^2+y^2+\frac{1}{4}-y \\
\Rightarrow & & 2 y & =0 \\
\Rightarrow & y & =0
\end{array}
$$
Locus of $z$ is $x$-axis.
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