Question

Let $z=a-\frac{i}{2};a\in R$. then $|i+z{|}^2-|i-z{|}^2=$

• A. 2
• B. -2
• C. 4
• D. -4

Answer 1

Answer: (B)

-2

$z=a-$ $\frac{i}{2}$

$|i+z{|}^2-|i-z{|}^2=|i+a-\frac{i}{2}{|}^2-{|i-(a-\frac{i}{2})|}^2$

${\left|\frac{2a+i}{2}\right|}^2-{\left|\frac{3i-2a}{2}\right|}^2$

Since $|z{|}^2=z.\overline{z}$

$\frac{1}{4}\left[\right(2a+i\left)\right(2a-i)-(-2a+3i\left)\right(-2a-3i\left)\right]$

$\frac{1}{4}\left[\right(4a^2-i^2)-((-2a{)}^2-(3i{)}^2\left)\right]$

$\frac{-8}{4}=-2$