Let $p$ be an odd prime, $k = (p-1)/2$, and $\zeta$ be a primitive $p$-th root of unity. The Fekete polynomial is defined to be \begin{equation*} f_p(t)= \sum_{a = 1}^{p-1} \left(\frac{a}{p} \right) t^a \end{equation*} where $\left(\frac{a}{p} \right)$ is the Legendre symbol. It is known that \begin{equation*} f_p(\zeta^k)= \epsilon_p\sqrt{p} \left(\frac{k}{p}\right) \end{equation*} where \begin{equation*} \epsilon_p = \begin{cases} 1 &\text{ if } p \equiv 1 \mod 4 \\ i &\text{ if } p \equiv 3 \mod 4 \end{cases} \end{equation*}
Now, if $p$ is a large prime congruent to $3$ mod $4$, then we have that $\zeta^k$ is close to $-1$ in the complex plane, and since $f_p(t)$ is continuous we also have that $f_p(\zeta^k) = \sqrt{-p}\left(\frac{k}{p}\right) \approx f_p(-1)$. But this seems to be impossible since $f_p(-1) \in \mathbb{Z}$ and $f_p(\zeta^k)$ is far up or down the imaginary axis. There must be something stupid that I'm missing here but I can't figure out what it is....