it vanishes on any curve in $\C$. Hence every curve in $\C$ is Lagrangian. For $N$ to be special Lagrangian with phase $e^{i\theta}$ we need that $$\operatorname{Re}(e^{-i\theta}\Upsilon)=\cos\theta\d x+\sin\theta\d y$$ is the volume form on $N$, or equivalently that $$\operatorname{Im}(e^{-i\theta}\Upsilon)=\cos\theta\d y-\sin\theta\d x$$ vanishes on $N$. This means that $\cos\theta\partial_x+\sin\theta\partial_y$ is everywhere a unit tangent vector to $N$, so $N$ is a straight line given by $N=\{(t\cos\theta ,t\sin\theta )\in\R^2\,:\,t\in\R\}$ (up to translation), so it makes an angle $\theta$ with the $x$-axis, hence motivating the term “phase $e^{i\theta}$”.

Notice that this result is compatible with the fact that special Lagrangians are minimal, and hence must be geodesics in $\R^2$; i.e. straight lines.

Consider $\C^2=\R^4$. We know that $\omega=\d x_1\wedge\d y_1+\d x_2\wedge\d y_2$. Since $\Upsilon=\d z_1\wedge \d z_2=(\d x_1+i\d y_1)\wedge (\d x_2+i\d y_2)$, we also know that $\operatorname{Re}\Upsilon=\d x_1\wedge\d x_2+\d y_2\wedge\d y_1$, which looks somewhat similar. In fact, if we let $J'$ denote the complex structure given by $J'(\partial_{x_1})=\partial_{x_2}$ and $J'(\partial_{y_2})=\partial_{y_1}$, then $\operatorname{Re}\Upsilon=\omega'$, the Kähler form corresponding the complex structure $J'$. Hence special Lagrangians in $\C^2$ are complex curves for a different complex structure.

In fact, we have a hyperkähler triple of complex structures $J_1,J_2,J_3$, where $J_1=J$ is the standard one and $J_3=J_1J_2=-J_2J_1$ so 