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Let $z=a-\frac{i}{2} ; a \in R$. Then $|i+z|^2-|i-z|^2$ is equal to
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-2
$\begin{aligned} & \text { Given, } z=a-\frac{i}{2} \\ & \begin{aligned} & \therefore|i+z|^2-|i-z|^2=\left|a+\frac{i}{2}\right|^2-\left|-a+\frac{3 i}{2}\right|^2 \\ &=a^2+\left(\frac{1}{2}\right)^2-\left(a^2+\left(\frac{3}{2}\right)^2\right) \\ &=\frac{1}{4}-\frac{9}{4}=-2\end{aligned}\end{aligned}$
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