On $\C^n$, $|\Upsilon(e_1,\ldots,e_n)|\leq 1$ for all unit vectors $e_1,\ldots,e_n$ with equality if and only if $P=\operatorname{Span}\{e_1,\ldots,e_n\}$ is a Lagrangian plane, i.e. $P$ is an $n$-plane such that $\omega|_P\equiv 0$.

Let $e_1,\ldots,e_n$ be the standard basis for $\R^n$ and let $P$ be an $n$-plane in $\C^n$. There exists $A\in\operatorname{GL}(n,\C)$ so that $f_1=Ae_1,\ldots,f_n=Ae_n$ is an orthonormal basis for $P$. Then $\Upsilon(Ae_1,\ldots,Ae_n)=\det_{\C}(A)$ so $$|\Upsilon(f_1,\ldots,f_n)|^2=|\operatorname{det}_{\C}(A)|^2=|\operatorname{det}_{\R}(A)|=|f_1\wedge Jf_1\wedge\ldots\wedge f_n\wedge J f_n|\leq |f_1||Jf_1|\ldots|f_n||Jf_n|=1$$ with equality if and only if $f_1,Jf_1,\ldots,f_n, Jf_n$ are orthonormal. However, this is exactly equivalent to the Lagrangian condition, since $\omega(u,v)=g(Ju,v)$ so $\omega|_P\equiv 0$ if and only if $JP=P^{\perp}$.

A submanifold $N$ of $M$ calibrated by $\operatorname{Re}(e^{-i\theta}\Upsilon)$ is called *special Lagrangian* with phase $e^{i\theta}$. If $\theta=0$ we say that $N$ is simply special Lagrangian. By Theorem \[Lagcalib.thm\], we see that $N$ is special Lagrangian if and only if $\omega|_N\equiv 0$ (i.e. $N$ is Lagrangian) and $\operatorname{Im}\Upsilon|_N\equiv 0$ (up to a choice of orientation so that $\operatorname{Re}\Upsilon|_N>0$).

Consider $\C=\R^2$ with coordinates $z=x+iy$, complex structure $J$ given by $Jw=iw$, Kähler form $\omega=\d x\wedge\d y=\frac{i}{2}\d z\wedge\d\overline{z}$ and holomorphic volume form $\Upsilon=\d z=\d x+i\d y$. We want to consider the special Lagrangians in $\C$, which are 1-dimensional submanifolds or curves $N$ in $\C=\R^2$.

Since $\omega$ is a 2-form, 