 Solution

A point is ramified with multiplicity $m_i$ exactly when it is a zero of $f^{\prime }$ of multiplicity $m_i-1$. Since $f^{\prime }$ has degree $d-1$, the fundamental theorem of algebra gives ${\sum }_i(m_i-1)=d-1$.

Conversely, choose distinct complex numbers $a_i$ and put $g(z)=c{\prod }_i(z-a_i{)}^{m_i-1}$ with $c\ne 0$. The assumption makes $g$ a polynomial of degree $d-1$. Any antiderivative $f$ of $g$ has degree $d$ and precisely the prescribed ramification multiplicities.