Question

The modulus of $\sqrt{2}i-\sqrt{-2}i$ is

Answer 1

Given $\sqrt{2i}-\sqrt{-2i}=\sqrt{2i}-\sqrt{2}i.$ $i[\because \sqrt{-1}=i]$

= $\sqrt{2}i+\sqrt{2}$ $[\because i^2=-1]$

= $\sqrt{2}+\sqrt{2}i$

= $\therefore |\sqrt{2}i-\sqrt{2}i|=|\sqrt{2}+\sqrt{2}i|$

= $\sqrt{(\sqrt{2}{)}^2+(\sqrt{2}{)}^2}$

= $\sqrt{2+2}$

= $\sqrt{4}$

= 2.