Find minimal polynomial for a period of 11th root of unity 
Let $\alpha=\zeta_{11}+\zeta_{11}^3+\zeta_{11}^4+\zeta_{11}^5+\zeta_{11}^9$. Find the minimal polynomial $m_{\alpha,\mathbb{Q}}(x)$, and likewise for $\gamma=\zeta_{11}+\zeta_{11}^{-1} $.

I started off by noting that since 11 is prime there is a $\beta$ such that: $\alpha+\beta=-1$, by the 11:th cylotomic polynomial. Here $\beta=\zeta_{11}^2+\zeta_{11}^6+\zeta_{11}^7+\zeta_{11}^8+\zeta_{11}^{10}$.
Also I have found $\alpha\beta=4+2(\alpha+\beta)$ and so $\alpha$ is a root of the polynomial: 
$$x^2+x+4+2(\alpha+\beta)=(x-\alpha)(x-\beta)$$
Is this correct? And if so how do we find the minimal polynomial for $\gamma$? 
I believe this has something to do with 
$$\mathbb{Q}(\zeta_{11})\cong(\mathbb{Z}|_{11}\mathbb{Z})^x\cong \mathbb{Z}|_{10}\mathbb{Z}$$
but I haven't made any advances as to how. 
 A: You result looks correct for $m_{\alpha,\mathbb{Q}}(x)$. Now you need to expand the powers of $\gamma$. I have chosen to align about the $\zeta^{-1}$ term so that addition in columns can be done with ease.
\begin{align*}
\gamma^5=\zeta^5+5\zeta^3+10\zeta+&10\zeta^{-1}+5\zeta^{-3}+\zeta^{-5}\\
-4\gamma^3=-4\zeta^3-12\zeta-&12\zeta^{-1}-4\zeta^{-3}\\
3\gamma=3\zeta+&3\zeta^{-1}
\end{align*}
and (aligned about the 6)
\begin{align*}
\gamma^4=\zeta^4+4\zeta^2+&6+4\zeta^{-2}+\zeta^{-4}\\
-3\gamma^2=-3\zeta^2-&6-3\zeta^{-2}
\end{align*}
It follows that 
\begin{equation}
\gamma^5+\gamma^4-4\gamma^3-3\gamma^2+3\gamma+1=0 \hspace{1em} (1)
\end{equation}
This gives you your minimal polynomial.
Edit:
We know that $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\cong (\mathbb{Z}/11\mathbb{Z})^\times$, so the Galois group has order 10. In addition, we have that 
\begin{equation*}
X^2-\gamma X+1\in\mathbb{Q}(\gamma)[X]
\end{equation*}
has $\zeta$ as a root. This tells us that $[\mathbb{Q}(\zeta):\mathbb{Q}(\gamma)]\leq2$. The relationship for gamma in (1) tells us that $[\mathbb{Q}(\gamma):\mathbb{Q}]\leq 5$. $\gamma$ must satisfy some algebraic relationship in a polynomial of degree less than or equal to 5. By order considerations,
\begin{equation*}
10=[\mathbb{Q}(\zeta):\mathbb{Q}]=[\mathbb{Q}(\zeta):\mathbb{Q}(\gamma)][\mathbb{Q}(\gamma):\mathbb{Q}]
\end{equation*}
the inequalities are equalities! In other words, since $[\mathbb{Q}(\gamma):\mathbb{Q}]=5$, I knew to look for a quintic relationship.
