Question

If $\alpha$ and $\beta$ are complex conjugates to each other and $\alpha =-\sqrt{2}+i$ then find ${\alpha }^2+{\beta }^2-\alpha \beta$.

Answer 1

Given $\alpha =-\sqrt{2}+i$

Since $\alpha$ and $\beta$ are complex conjugates to each other we get $\beta =-\sqrt{2}-i$

Now we have to find ${\alpha }^2+{\beta }^2-\alpha \beta$

${\alpha }^2+{\beta }^2-\alpha \beta =(-\sqrt{2}+i{)}^2+(-\sqrt{2}-i{)}^2-(-\sqrt{2}+i)(-\sqrt{2}-i)$

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

$=2-1+2-1-(2+1)[\because i^2=-1]$

$=2-3$

$=-1$

Thus ${\alpha }^2+{\beta }^2-\alpha \beta =-1$