Cotangent bundle of complex manifold is Calabi-Yau manifold We say that a complex manifold $M$ is Calabi-Yau if the canonical bunlde is trivial $K_M=0$. How can we prove that the total space of the cotangent bundle of a compact complex manifold $N$ is Calabi-Yau $2n$-fold, where $n$ is the dimension of $N$? 
 A: Here's a more general statement from an algebraic geometry standpoint -- I do not really know complex geometry but I'd expect the statements to translate wel.  Let $X$ be a projective variety and $\mathcal{E}$ a locally free sheaf.  Recall that we can define the total space as $E = \mathcal{Spec}_{\mathcal{O}_X} \mathcal{Sym}_{\mathcal{O}_X} \mathcal{E}^\vee$ and there is a natural map $\pi: E \rightarrow X$.
I want to express the canonical bundle $\omega_E$ in terms of $\mathcal{E}$ and $\omega_X$:
$$\omega_E = \pi^* \omega_X \otimes \pi^* \bigwedge^{top} \mathcal{E}^\vee$$
To realize this, I need a result like: every locally free sheaf is a subsheaf of a sum of trivial locally free sheaves.  This gives us a short exact sequence of (fibers of) vector bundles:
$$0 \rightarrow E \rightarrow V \rightarrow T \rightarrow 0$$
where $V$ is a trivial bundle, which corresponds to a short exact sequence of locally free sheaves
$$0 \rightarrow \mathcal{E} \rightarrow \mathcal{V} \rightarrow \mathcal{T} \rightarrow 0$$
Now, notice that $\mathcal{T}^\vee$ is locally the ideal sheaf cutting out $E$ in $V$.  Thus tensoring with $\mathcal{O}_E$ gives
$$\mathcal{N}^\vee = \mathcal{Sym}_{\mathcal{O}_X}\mathcal{T}^\vee$$
And now, using the conormal short exact sequence for differentials and taking the top exterior power, we have (abusing notation, using a correspondence between line bundles on $X$ and line bundles on any vector bundle of $X$):
$$\omega_E = \omega_X \otimes \bigwedge^{top} \mathcal{T}$$
Finally use the original short exact sequence in locally free sheaves and take the top exterior powers to find that $\bigwedge^{top}\mathcal{T} = \bigwedge^{top}\mathcal{E}^\vee$ so the result follows.
Now, for your question, $\mathcal{E}$ is the locally free sheaf of differentials, so the dual of its top exterior power is $\omega^\vee$, and $\omega^\vee \otimes \omega = \mathcal{O}$ and that's it.
