If a Jacobian has complex eigenvalues, must the real part equal the trace?

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My course notes (BSc Mathematics, second year module in ordinary differential equations, unpublished) tell me the following (I paraphrase).

Consider a 2 x 2 Jacobian $J$ for a fixed point $P$. Suppose the eigenvalues $\lambda_1,\lambda_2$ are complex. Then they are a pair of complex conjugates, $\sigma+i\omega,\sigma-i\omega$, where $\sigma,\omega\in\mathbb{R}$, and $\sigma=\text{Tr}(J). \tag{1}$

But one of the homework exercises seems to contain a counterexample.

Let $J=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$. Then, according to my working, the lecturer's worked solution, and this online calculator, the eigenvalues are $\lambda_1,=1-i,\lambda_2=1+i$. But then $\sigma=1\neq2=1+1=\text{Tr}(J)$, so $(1)$ doesn't hold.

What's gone wrong?