The complex tangent bundle of $\mathbb{C}P^1$ is not isomorphic to its conjugate bundle. In " Characteristic Classes" by Milnor and Stasheff on pages 167-168, the authors give a brief argument about why: 
The complex tangent bundle of $\mathbb{C}P^1$ is not isomorphic to its conjugate bundle.
However I don't understand their argument. I think I understand that if we did have an isomorphism between these two bundles then it necessarily swaps the orientations on the fibers in the associated real bundles, thus in every fiber there is a fixed line that is invariant under the isomorphism. I don't understand the next sentence " ...the 2 sphere does not admit any continuous field of tangent lines." and why it implies what we are trying to prove. I think maybe they are trying to show that this implies that the tangent bundle splits and hence the Euler class is 0. But we know that it is nonzero so we have a contradiction. But I am pretty confused. 
 A: It is easy to show that an isomorphism of conjugate complex structures on a plane must be a reflection. 
The axes of reflection determine a line field on the sphere.
If the field is orientable then it will have a non zero section and the sphere would have Euler characteristic zero.
If it is non-orientab;e then it will determine a on-zero first Z2 cohomology class,the Stiefel-Whitney class of the line bundle. But the first cohomology of the 2 sphere is zero.
A: This may be the blind leading the blind, but I believe your argument is right. 
The authors might also refer to their argument on p. 101, where they say that if the tangent bundle $TM$ of a smooth manifold is oriented and $e(TM) \neq 0$, then $TM$ can't admit any odd-dimensional subbundle $\xi$. This may be due to Property 9.4; we would have $2e(\xi) = 0$ and thus $2e(TM) = 2e(\xi) \smile e(\xi^\perp)= 0$, despite it being the case that $H^{\dim M}(M) \cong \mathbb{Z}$. 
A related thing to think about is that, given a continuous line field $\xi$, picking a unit vector in each line would give a nowhere-vanishing vector field $v$ on $S^2$, which would contradict the hairy ball theorem. You'd need to show that given a continuous line bundle, you could continuously pick basis vectors, but I think that would follow from orientability.
Also, rotating each vector $v_p \in T_p S^2$ by $\pi/2$ would give another nowhere-vanishing vector field $w$, and together, $v$ and $w$ would parallelize the tangent bundle $TS^2$, which is impossible (for example because the Euler class is nonzero).
