Question

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

Answer 1

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

${|i+z|}^2-{|i-z|}^2$

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

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

$[\because z=x+iythen|z|=\sqrt{x^2+y^2}and|z{|}^2=x^2+y^2]$

Since $a$ is real number than:

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

$\Rightarrow \frac{1}{4}-\frac{9}{4}=-2$

${|i+z|}^2-{|i-z|}^2=-2$

Answer.