Let z = a - frac{i}{2},where a in R. Then |i | vedclass

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(B) Given,$z = a - \frac{i}{2}$.We need to calculate $|i + z|^2 - |i - z|^2$.First,substitute $z$ into the expressions:$i + z = i + a - \frac{i}{2} =

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$-2$

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(B) Given,$z = a - \frac{i}{2}$.We need to calculate $|i + z|^2 - |i - z|^2$.First,substitute $z$ into the expressions:$i + z = i + a - \frac{i}{2} = a + \frac{i}{2}$.$i - z = i - (a - \frac{i}{2}) = -a + \frac{3i}{2}$.Now,calculate the squared moduli:$|i + z|^2 = |a + \frac{i}{2}|^2 = a^2 + (\frac{1}{2})^2 = a^2 + \frac{1}{4}$.$|i - z|^2 = |-a + \frac{3i}{2}|^2 = (-a)^2 + (\frac{3}{2})^2 = a^2 + \frac{9}{4}$.Finally,subtract the two values:$|i + z|^2 - |i - z|^2 = (a^2 + \frac{1}{4}) - (a^2 + \frac{9}{4}) = \frac{1}{4} - \frac{9}{4} = -\frac{8}{4} = -2$.

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

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