Let $\xi_{19} \in G_{19}$ be a primitive root of unity. Find $\Re\sum_{k=1}^9\xi_{19}^{k^2}$. Question

Let $\xi_{19} \in G_{19}$ be a primitive root of unity. Find $\Re\sum_{k=1}^9\xi_{19}^{k^2}$.

Attempt
I'm having doubts about how I'm solving this exercise, this is what I did:
$$
\Re\left(\sum_{k=1}^9\xi_{19}^{k^2}\right)=\sum_{k=1}^9\Re\left(\xi_{19}^{k^2}\right)\\
=\Re\xi^1+\Re\xi^4+\Re\xi^9+\Re\xi^{16}+\Re\xi^{25}+\Re\xi^{36}+\Re\xi^{49}+\Re\xi^{64}+\Re\xi^{81}\\
=\Re\xi^1+\Re\xi^4+\Re\xi^9+\Re\xi^{-3}+\Re\xi^{6}+\Re\xi^{-2}+\Re\xi^{-8}+\Re\xi^{7}+\Re\xi^{5}\\
=\Re\xi^1+\Re\xi^4+\Re\xi^9+\Re (\overline\xi)^{3}+\Re\xi^{6}+\Re(\overline\xi)^{2}+\Re(\overline \xi)^{8}+\Re\xi^{7}+\Re\xi^{5}
$$
Also, using $\Re(z)=\Re(\overline z)$, the sum is then simplified to:
$$\Re \sum_{k=1}^{9}\xi^i=\Re \frac {\xi^{10}-\xi}{\xi-1} .$$
Is what I did thus far correct? Can this be simplified further?

I'm not exactly sure how exactly you got rid of the squares in the exponents. That step definitely lacks justification and is probably incorrect.
Here's what you can do:
Let $A = \sum_{k=1}^{9}\xi_{19}^{k^2}$ and $B = \sum_{k=1}^{9}\xi_{19}^{-k^2}$. Note that the exponents in $A$ and $B$ are precisely the sets of all squares and non-squares mod $19$. (Excluding $0$.) Then we can verify that
$$ AB = 9+4\sum_{k=0}^{18}\xi_{19}^{k} = 5,$$
by verifying that the pairwise sums of all squares and non-squares in $\mathbb F_{19}$ will give every $x\neq 0$ precisely $4$ times and $0$ precisely $9$ times.(*)
On the other hand $1+A+B = \sum_{k=0}^{18}\xi_{19}^{k} = 0$. So we may solve the system for $A$ and $B$. We find
$$ A,B = \frac{-1\pm\sqrt{-19}}{2}, $$
and thus $\Re A=\Re B=-1/2$. 

(*) I have checked this fact using the programming language Python, but it is probably not too hard to check it by hand. 

L1 = sorted([(i^2)%19 for i in range(1,10)]])
  L2 = sorted([(-i^2)%19 for i in range(1,10)])
  sorted([(i+j)%19 for i in L1 for j in L2])

