TITLE: Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$
QUESTION [15 upvotes]: Let $\alpha$, $\beta$, $\gamma$, $\delta$ be the roots of $$z^4-2z^3+z^2+z-7=0$$ then find value of $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$$
Are Vieta's formulas appropriate?

REPLY [46 votes]: There is no need to use Vieta's formulas. Let 
$$f(z)=z^4-2z^3+z^2+z-7=(z-\alpha)(z-\beta)(z-\gamma)(z-\delta).$$
Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that
$$(\alpha^2+1)(\beta ^2+1)(\gamma^2+1)(\delta^2+1)=f(i)f(-i)=|f(i)|^2=|-7+3i|^2=49+9=58.$$