If $p>0$ demonstrate that the $1/2\pi i$ the line integral of $z^p f'(z)/f(z)$ is $\sum (z_k)^p$

Question

This is basically a deviation of Rouche's Theorem from what I can tell. My first instinct was to do this via induction in which we know that $p=0$ we would have Rouche's theorem. But it gets trickier when you try to do it for cases like $p+1$ for instance. Then it varies more or less depending on $f(z)$. Can any of you give me some suggestion on tackling this?

Answer 1

Let $f,g: D \to \mathbb{C}$ be analytic functions. Then one can show that if $\gamma$ is a simple closed curve in $D$ enclosing finitely many roots $z_k$ of $f$, then $$ \int_{\gamma} g(z) \frac{f'(z)}{f(z)} \, dz = 2\pi i\sum_{k} n_k g(z_k), $$ where $n_k$ is the order of the root $z_k$.

Proof: the integrand is meromorphic, with singularities at $z_k$, and nowhere else. By the Residue Theorem, it suffices to compute the residues at $z_k$, sum them, and multiply by $2\pi i$. By the definition of analytic, one can find for $z_k$ a sufficiently small disc $D_k$ on which $f(z) = (z-z_k)^{n_k} h(z)$, where $h(z)$ is an analytic function that does not vanish on $D_k$. Then the integrand can be evaluated in this disc as $$ g(z) \frac{f'(z)}{f(z)} = g(z) \frac{n_k (z-z_k)^{n_k-1}h(z) + (z-z_k)^{n_k}h'(z)}{(z-z_k)^{n_k}h(z)} = g(z) \left( \frac{n_k}{z-z_k} + \frac{h'(z)}{h(z)} \right), $$ and hence, since $h(z)$ does not vanish, the residue at $z_k$ is $n_k g(z_k)$, and the Residue Theorem gives you the rest. $\square$

---

Remark: there are generalisations of this to meromorphic $f$, but it then becomes more complicated to express: one uses the definition of pole instead of zero, and basically everything else comes out in the same way.