I am starting learning complex analysis by myself. Here is the question I am trying to solve and its solution:
But I do not understand why the antiderivative contains 2? and why we require the condition that "any simple closed curve does not pass through the origin" ?
Could someone explain this to me please?
The $2$ is a mistake: the antiderivative is just $-1/z$. The curve should not pass through the origin because $z=0$ is a singularity of $1/z^k$.
EDIT: Here is my solution to the question.
$$(z-w)^{2m} = \sum_{j=0}^{2m} {2m \choose j} z^j (-w)^{2m-j} $$ so by the Residue Theorem, if $C$ is a positively-oriented circle about the origin, and $2m \ge n-1$, $$ \eqalign{0 &= \frac{1}{2\pi i}\oint_C f(z) (z-w)^{2m} \; dz = \sum_{k=1}^n \frac{c_k}{2\pi i}\oint_C z^{-k} (z-w)^{2m} \; dz\cr &= \sum_{k=1}^n c_k {2m \choose k-1} (-w)^{2m-k+1}} $$ This is a polynomial in $w$ that is identically $0$, so its coefficients are all identically $0$, i.e. all $c_k = 0$.