Show that if \(f\) is holomorphic in \(\DD_r(a)\) and \(a\) is a zero of \(f\) of multiplicity \(n\), then \(f^{(k)}(a) = 0\) for \(k\leq n-1\) and \(f^{(n)}(a) \neq 0\). Show that this is an iff.
Solution
Write \(f(z) = \sum_{k\geq 0} c_k (z-a)^k\) on \(\DD_r(a)\), where \(c_k = f^{(k)}(a)/k!\). Recall that \(a\) is a zero of multiplicity \(m\) when \(f(z)=(z-a)^mg(z)\) with \(g\) holomorphic near \(a\) and \(g(a)\neq0\).
\(\impliedby\): Suppose \(f^{(k)}(a)=0\) for \(k\le m-1\) and \(f^{(m)}(a)\ne0\), so \(c_k=0\) for \(k\le m-1\) and \(c_m\ne0\). Then \[ f(z) = \sum_{k\geq m} c_k (z-a)^k = (z-a)^m \sum_{k\geq m} c_k (z-a)^{k-m} \da (z-a)^m g(z) ,\] where \(g\) is holomorphic on \(\DD_r(a)\) with \(g(a) = c_m \neq 0\), making \(a\) a zero of \(f\) of multiplicity \(m\).
\(\implies\): If \(a\) is a zero of multiplicity \(m\), write \(f(z) = (z-a)^m h(z)\) with \(h\) holomorphic near \(a\) and \(h(a)\neq0\), and expand \(h(z)=\sum_{j\ge0}d_j(z-a)^j\) with \(d_0=h(a)\). Then \(f(z)=\sum_{j\geq0}d_j(z-a)^{j+m}\), and by uniqueness of power series coefficients, \(c_k=0\) for \(k\le m-1\) and \(c_m=d_0\neq0\). That is, \(f^{(k)}(a)=0\) for \(k\le m-1\) and \(f^{(m)}(a)=m!\,h(a)\neq0\).