Let $M$ be a complex 2-dimensional manifold. I am suggested to prove that the Todd class of $TM$ is $1$, but i can't quite believe that it indeed holds.
Given a bundle $\xi: E \rightarrow X$ of rank $n$ we define its $k$-th Chern class as the the coefficient that corresponds to the $k$-th monomial in the following decomposition:
$$ \text{det} (I + \frac{i t \Omega}{2 \pi}) = \sum_{k}{c_{k}(X) t^{k}}$$ where $\Omega$ is the curvature 2-form associated with the canonical connection on $X$.
According to the splitting principle, $\xi$ breaks into the sum of line bundles $$\xi = L_{1} \oplus L_{2} \oplus \ldots L_{n}$$ Define $x_{i} = c_{1}(L_{i})$. Define the Todd class of $M$ $$ \text{Td}(M) = \prod_{i = 1}^{n}{\frac{x_{i}}{1 - e^{-x_{i}}}}$$ where $x_{i}$ are the Chern roots of the tangent bundle $TM \rightarrow M$.
After some computations with the Chern root sone can show that $$\text{Td}(M) = 1 + \frac{c_{1}}{2} + \frac{c_{1}^{2} + c_{2}}{12} + \ldots $$
(Note that here $\text{Td}(M)$ stands for the complete Todd class that is the sum $\text{Td}(M) = 1 + \text{Td}_{1}(M) + \ldots$)
the first argument Assume that for a bundle of rank $2$ the Todd class vanishes, then all Chern classes vanishes as well, but this certainly does not holds, since of course there are two-dimensional manifolds whose tangent bundle is non-trivial.
Also, using the Riemann-Roch formula one can deduce the following formula for the Todd classes $$\int_{M}{\text{Td}_{n}(M)} = \sum_{k = 0}^{n} {\text{dim}{H^{k}(X, \mathcal{O_{X}})}}$$ where $\mathcal{O}_{X}$ is the structure sheaf of $X$. Assume that the first Todd class is $1$, then the left hand side integrates to $1$ (since we assume that volume form is normalized), whilst the right hand side is the Euler characteristic which fails to be equal $1$ for every 2-dimensional manifold.
I have a feeling that i'm missing something important here. Is it true that the Todd class of a 2-dimensional manifold is 1? If not, why do the mentioned reasonings fail to be correct?