in a probelm set I found $(x + \sqrt2j)(x − \sqrt2ij) =(x^2 + 2x + 2)$ where $j=√i$
but i can't understand how this happened. I have done this $(x + \sqrt2j)(x − \sqrt2ij)=x^2-2i^2+x\sqrt2j(1-i)=x^2-2+x\sqrt2j(1-i)$
then what will happen please help me
First of all, note that $-2i^2 = 2$, not $-2$. Second, we do in fact have $$ \sqrt{2}\,j(1 - i) = 2 $$ so your result makes sense. That is, with careful analysis, we find that $$ x^2-2i^2+x\sqrt2j(1-i) = x^2 + 2x + 2 $$ In fact, instead of using "$j$", note that $$ \sqrt{i} = [\cos(\pi/2) + i\sin(\pi/2)]^{1/2} = \cos(\pi/4) + i\sin(\pi/4) = \frac{1}{\sqrt{2}}(1 + i) $$