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

Attempt

For $\rho$ small enough, we have:

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

and

\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

\end{align}

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