Take the equation $y^{d}=\Pi\_{1}^{n}(x-t\_{i})^{m\_{i}}$ over $\mathbb{C}$. This affine equation gives a cyclic cover of $\mathbb{P}^{1}$. Now it is usually said without explanation that if the sum $\sum m\_{i}$ is not divisible by $d$, then the family is ramified over $\infty$. My question is how to see explicitly--in terms of equations maybe--that this is the case. 



How to explicitly see the ramification over infinity
One way to see this is the following. Let $U$ be the complement of the points $t\_1,\ldots,t\_n$ in $\mathbb{P}^1(\mathbb{C})$. The fundamental group $\pi\_1(U)$ is generated by elements $\gamma\_1,\ldots,\gamma\_n,\gamma\_\infty$ with relation 
$\gamma\_\infty^{-1} = \gamma\_1\cdots \gamma\_n$. Consider the character 
$\chi:\pi\_1(U)\to \mathbb{C}^\*$ defined by 
$\chi(\gamma\_i) = e^{2\pi m\_i/d}, i = 1,\ldots,n$, 
$\chi(\gamma\_\infty) = e^{-2\pi M/d}$, where $M = \sum m\_i$. Let $V$ be the quotient of the universal cover of $U$ by the kernel of $\chi$.
The normalization of your curve is isomorphic to the normalization of $\mathbb{P}^1$ in $V$. It is a smooth complete curve on which the cyclic group of order $d$ acts as the Galois group with quotient $\mathbb{P}^1$. The ramification over $\infty$ is defined by the value of $\chi$ at $\gamma\_\infty$, i.e. $e^{2\pi M/d}$. It is trivial, and hence the cover is unramified over $\infty$ if and only if $d|M$. The ramification index is equal M\mod d.$


