If z = x + iy is a complex number satisfying | vedclass

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(D) Given the equation $\left| z + \frac{i}{2} \right|^2 = \left| z - \frac{i}{2} \right|^2$. Substitute $z = x + iy$: $\left| x + iy + \frac{i}{2} \r

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$x$-axis

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(D) Given the equation $\left| z + \frac{i}{2} \right|^2 = \left| z - \frac{i}{2} \right|^2$. Substitute $z = x + iy$: $\left| x + iy + \frac{i}{2} \right|^2 = \left| x + iy - \frac{i}{2} \right|^2$ $\left| x + i(y + \frac{1}{2}) \right|^2 = \left| x + i(y - \frac{1}{2}) \right|^2$ Using the property $|a + ib|^2 = a^2 + b^2$: $x^2 + (y + \frac{1}{2})^2 = x^2 + (y - \frac{1}{2})^2$ $x^2 + y^2 + y + \frac{1}{4} = x^2 + y^2 - y + \frac{1}{4}$ $y = -y$ $2y = 0$ $y = 0$ This represents the $x$-axis.

If $z = x + iy$ 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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