Residue identity for function composition: $ \operatorname{Res}(f; h(a)) = \operatorname{Res}((f∘h)h'; a)$

2018-06-24 11:00:23

With $a \in \mathbb{C}$, let $h: D_1 \to D_2$ and $f: D_2\setminus{\{h(a)\}} \to \mathbb{C}$ be analytic functions. Also require $h'(a) \neq 0$.

Claim: $$ \operatorname{Res}(f; h(a)) = \operatorname{Res}((f \circ h)h'; a)$$


For $\rho$ small enough, we have:

$$ \operatorname{Res}(f; h(a)) = \frac{1}{2\pi i}\int_{|u-h(a)|=\rho} f(u)\ \mathrm{d}u$$


\begin{align}\operatorname{Res}((f \circ h)h', a)& = \frac{1}{2\pi i} \int_{|u-a|=\rho} (f \circ h) h'\ \mathrm{d}u = \frac{1}{2\pi i} \int_0^{2\pi} f(h(a + \rho e^{it})) h'(a + \rho e^{it})\rho e^{it} i\ \mathrm{d}t\\&= \frac{1}{2\pi i} \int_\alpha f(u)\ \mathrm{d}u


where $\alpha: [0, 2\pi] \to \mathbb{C}, \alpha(t) = h(a+\rho e^{it})$.

We see that the curve of the LHS defined by $|u-h(a)|=\rho$ (i.e. $h(a) + \rho e^{it}$) is "approximated" by the curve defined by $\alpha$ for $\rho \to 0$ because analytic functions approximately retain circles (see here).

(Note that we already used $h'(a) \neq

  • Let $U \subset D_1$ be a disk with center $a$ in which $h$ is injective.

    Then $V := h(U) \subset D_2$ is a simply-connected neighbourhood of $h(a)$.

    For $\rho > 0$ small enough, let


    \gamma: [0, 2 \pi] \to U, \gamma(t) = a + \rho e^{it}


    and $\alpha = h \circ \gamma$. Then – as you already computed –


    \operatorname{Res}((f \circ h)h', a) = \frac{1}{2 \pi i}\int_\gamma f(h(z))h'(z) \, dz = \frac{1}{2 \pi i}\int_\alpha f(w) \, dw \quad .


    Since $V$ is simply-connected and $f$ is holomorphic in $V \setminus \{ h(a) \}$, we can apply the Residue theorem to

    the right-hand side:


    \frac{1}{2 \pi i}\int_\alpha f(w) \, dw = \operatorname{Res}(f, h(a)) I(\alpha, h(a))


    and it remains to show that the winding number $I(\alpha, h(a))$ of $\alpha$ with respect to $h(a)$ is one:


    I(\alpha, h(a)) = \frac{1}{2 \pi i}\int_\alpha \frac{dw}{w - h(a)}

    = \frac{1}{2 \pi i}\int_\gamma \frac{h'(z)}{h(z) - h(a)} \, dz

    = \operatorname{Res}(\frac{h'}{h-a}, a)


    because $ h'/(h-a)$ is holo

    2018-06-24 13:16:18