I posted a question earlier. This question is smaller, more to the point and more basic I hope. Yet it is at the heart of my problems, I think.

In one of my references, degree k meromorphic differentials with poles of order $k$ at $n$ points $z_i$ on the Riemann sphere are mentioned: $\phi_k(z)=F(z)dz^k$. Here $F(z)$ is a rational function on the complex plane.

If we want to know the behaviour of $\phi_k(z)$ at $\infty$ we change coordinates to $z'=1/z$ under which the k-differential transforms as $\phi_k(z')=F(1/z')(-dz'/(z')^2)^k$. Furthermore, these k-differentials are required to have fixed residues $\alpha_i$ on each $z_i$.

Then it is stated that we thus allow: $$\phi_k(z) \sim \frac{\alpha_i}{(z-z_i)^k} dz^k +... $$

I have two questions about this:

1) May I also think of these k-differentials as $$\phi_k(z)=\frac{\prod^m_{i=1}c(z-u_i)}{\prod^n_{i=1}(z-z_i)^k}dz^k$$ Here, $c$ is a constant, $m=n-2$ in order that the degree of the divisor of this k-differential is $-2k$ as required on a Riemann surface of genus $g=0$. Furthermore, some of the $u_i$ and/or $c$ may depend on the fixed residues $\alpha_i$. If this is wrong, what is meant by the above stated form of the k-differential?

2) What does it mean for a k-differential to have a residue? For instance, how does Cauchy's formula work for a quadratic differential? Does a second order pole behave for a quadratic differential as a simple pole behaves for a 1-differential? If so, I don't think the residues of a quadratic differential sum to zero.