Here's a baby example I'm working on:
Consider the manifold $\mathbb{R}^2$ and the symplectic form $dp\wedge dx$ where clearly the space is parameterized by $(x,p)$. For fixed $\alpha > 0$ consider the coordinate $z = x -i\alpha p$ and the corresponding tangent vector $$ \frac{\partial}{\partial z} \;\; =\;\; \frac{1}{2}\left (\frac{\partial}{\partial x} + \frac{i}{\alpha}\frac{\partial}{\partial p} \right ). $$ Consider the polarization $P$ on $\mathbb{R}^2$ given by the span of the vector $\partial_z$. This gives rise to a complex polarization (i.e. $P_z \cap \overline{P_z} = \{0\}$ for all $z \in \mathbb{R}^2$). Given the mapping $J:T_z\mathbb{R}^2 \to T_z\mathbb{R}^2$ where $J_z(\partial_z) = i\partial_z$ and $J_z(\partial_{\overline{z}}) = -i\partial_{\overline{z}}$, show that $\omega(X,JY)$ is positive-definite.
I've tried showing this a couple of different ways but to no avail. I'm pretty new to symplectic geometry and I'm pretty sure there's just some small detail I'm missing.
If I complexify the whole problem, we can see that the isomorphism under $\omega$ shows us that $dp = -(dz + d\overline{z})$ and $dx = -i\alpha(dz - d\overline{z})$ and therefore $$ \omega \;\; \equiv \;\; dp\wedge dx \;\; =\;\; 2i\alpha (d\overline{z}\wedge dz). $$
However taking a vector $X = a\frac{\partial}{\partial z} + b \frac{\partial}{\partial \overline{z}}$ we can easily compute that
\begin{eqnarray*} \omega(X,JX) & = & \omega \left (a\partial_z + b \partial_{\overline{z}}, ia\partial_z - ib \partial_{\overline{z}}\right ) \\ & = & 2iab \omega\left (\partial_{\overline{z}}, \partial_z \right ) \\ & = & -4\alpha ab. \end{eqnarray*}
I don't see how positive-definiteness of the form can be discerned here. We would need $a$ and $b$ to be different signs, let alone nonzero. I have also carried out the computation sticking with the coordinates $(x,p)$, but I get a similar issue when I do this (I'm omitting this from my question because I don't think it's any more illuminating than what I have written above). What am I missing here?
For reference, this is a question from Brian Hall's Quantum Theory for Mathematicians. It's in the final chapter discussing quantization of observables for manifolds.
If $z = x-i\alpha p$ then $dx = \tfrac{1}{2}(dz + d\bar{z})$ and $dp = \tfrac{i}{2\alpha}(dz - d\bar{z})$ so $\omega = dp \wedge dx = \tfrac{i}{2\alpha} dz \wedge d\bar{z}$.
You wrote $X = a\partial_z + b\partial_{\bar{z}}$ and $JX=ia\partial_z-ib\partial_{\bar{z}}$, and this gives $\omega(X,JX) = \tfrac{1}{\alpha}ab$.
But remember the $(x,p)$ coordinates: $\partial_x = \partial_z + \partial_{\bar{z}}$ and $\partial_p = -i\alpha(\partial_z - \partial_{\bar{z}})$, so $$X = A\partial_x + B\partial_p = (A-i\alpha B)\partial_z + (A+i\alpha B)\partial_{\bar{z}}$$ meaning $a = A-i\alpha B$ and $b = A+i\alpha B$, so we get $$\omega(X,JX) = \tfrac{1}{\alpha}(A-i\alpha B)(A+i \alpha B) = \tfrac{1}{\alpha}(A^2 + \alpha^2 B^2)$$ which is positive-definite.