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

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(A) We have,$\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$. Substituting $z=x+iy$: $\left|x+i(y+\frac{1}{2})\right|^2 = \left|x+i(y-\frac

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$x$-axis

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(A) We have,$\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$. Substituting $z=x+iy$: $\left|x+i(y+\frac{1}{2})\right|^2 = \left|x+i(y-\frac{1}{2})\right|^2$. Using the property $|a+ib|^2 = a^2+b^2$: $x^2+(y+\frac{1}{2})^2 = x^2+(y-\frac{1}{2})^2$. $x^2+y^2+y+\frac{1}{4} = x^2+y^2-y+\frac{1}{4}$. $y = -y$ $\Rightarrow 2y = 0$ $\Rightarrow y=0$. Since $y=0$,the locus of $z$ is 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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