Show that if \(z_0\) is a pole of order \(n\) of \(f\), then it is a pole of order \(n+k\) for \(f^{(k)}\).
Solution
Without loss of generality suppose \(z_0=0\) is the pole. On a punctured disk about \(0\), write the Laurent series \(f(z) = \sum_{j\geq -n} c_j z^j\) with \(c_{-n}\neq0\). Differentiating term by term, \[ f^{(k)}(z) = \sum_{j\geq -n} j(j-1)\cdots(j-k+1)\, c_j z^{j-k} .\] The most negative power that occurs is \(z^{-n-k}\), with coefficient \[ (-n)(-n-1)\cdots(-n-k+1)\,c_{-n}=(-1)^k n(n+1)\cdots(n+k-1)\,c_{-n}\neq0 ,\] so \(0\) is a pole of order \(n+k\) of \(f^{(k)}\).