Recover a quartic and deduce reciprocal roots

The coefficients are real, so $1-i$ is also a root. Form the conjugate quadratic and let the other factor be $F(z)=2z^2+Pz+Q$:

$\begin{aligned}2z^4+az^3+7z^2+bz+2&=[z-(1+i)][z-(1-i)]F(z)\\&=[(z-1)-i][(z-1)+i]F(z)\\&=[(z-1)^2-i^2](2z^2+Pz+Q)\\&=(z^2-2z+2)(2z^2+Pz+Q).\end{aligned}$

Compare the constant coefficient: $2Q=2$, so $Q=1$. The $z^2$ coefficient gives $1-2P+4=7$, hence $P=-1$. The $z^3$ coefficient gives $a=P-4=-5$. The $z$ coefficient is

$\begin{aligned}&b\\ & =2P-2Q\\ & =2(-1)-2(1)\\ & =-4.\end{aligned}$

Thus the factorisation is $(z^2-2z+2)(2z^2-z+1)$. The coefficient-of-$z$ line printed in the notes has an error; the product above gives the corrected comparison.

The remaining quadratic gives

$\begin{aligned}2z^2-z+1&=0,\\z&=\frac{1\pm\sqrt{1-4(2)(1)}}{2(2)}\\&=\frac{1\pm\sqrt7\,i}{4}.\end{aligned}$

Together with $1-i$, these are the roots other than the given $1+i$.

The constant term is 2, so neither polynomial has zero as a root. Divide the reversed equation by $z^4$:

\end{aligned}$$ Set $u=1/z$. Reordering gives $2u^4+au^3+7u^2+bu+2=0$, the original quartic. Therefore its four roots give the four reciprocals: $$\begin{aligned}\frac1{1+i}&=\frac1{1+i}\frac{1-i}{1-i}=\frac{1-i}{1^2+1^2}=\frac{1-i}2,\\\frac1{1-i}&=\frac1{1-i}\frac{1+i}{1+i}=\frac{1+i}{1^2+1^2}=\frac{1+i}2,\\\frac4{1+\sqrt7\,i}&=\frac4{1+\sqrt7\,i}\frac{1-\sqrt7\,i}{1-\sqrt7\,i}=\frac{4(1-\sqrt7\,i)}{1^2+(\sqrt7)^2}=\frac{1-\sqrt7\,i}2,\\\frac4{1-\sqrt7\,i}&=\frac4{1-\sqrt7\,i}\frac{1+\sqrt7\,i}{1+\sqrt7\,i}=\frac{4(1+\sqrt7\,i)}{1^2+(\sqrt7)^2}=\frac{1+\sqrt7\,i}2. \end{aligned}$$