[proofplan] We shrink the neighbourhood of $b$ so that $\varphi$ is a local holomorphic coordinate and so that $\varphi(w)-a$ has a simple zero at $w=b$. We then write the Laurent expansion of $f$ at $a$ and pull each Laurent term back under $z=\varphi(w)$. The coefficient with exponent $-1$ contributes exactly the same residue, while all other terms have zero residue: nonnegative powers are holomorphic, and negative powers other than $-1$ are derivatives of meromorphic functions. [/proofplan] [step:Shrink to a local holomorphic coordinate at $b$] By the [holomorphic inverse function theorem](/theorems/4950) (citing a result not yet in the wiki: Holomorphic [Inverse Function Theorem](/theorems/51)), since $\varphi'(b)\neq 0$, there exists an open neighbourhood $V_0 \subset V$ of $b$ such that $\varphi|_{V_0}: V_0 \to \varphi(V_0)$ is biholomorphic onto an open neighbourhood of $a$. Shrinking $V_0$ further if necessary, assume $\varphi(V_0)\subset U$. Define the [holomorphic function](/page/Holomorphic%20Function) \begin{align*} t: V_0 \to \mathbb{C}, \qquad w \mapsto \varphi(w)-a. \end{align*} Then $t(b)=0$ and $t'(b)=\varphi'(b)\neq 0$, so $t$ has a simple zero at $b$. Hence there is a holomorphic function \begin{align*} u: V_0 \to \mathbb{C} \end{align*} with $u(b)\neq 0$ and \begin{align*} t(w)=(w-b)u(w) \end{align*} for all $w\in V_0$. [/step] [step:Expand the meromorphic function at $a$] Since $f:U\to\mathbb{C}$ is meromorphic near $a$, its Laurent expansion at $a$ has finite principal part. Thus there exist an integer $N\geq 0$, complex coefficients $c_m\in\mathbb{C}$ for every integer $m\geq -N$, and an open punctured neighbourhood of $a$ on which \begin{align*} f(z)=\sum_{m=-N}^{\infty} c_m (z-a)^m. \end{align*} By the definition of the residue as the coefficient of $(z-a)^{-1}$ in the Laurent expansion of the meromorphic differential $f(z)\,dz$, we have \begin{align*} \operatorname{Res}_{z=a}\left(f(z)\,dz\right)=c_{-1}, \end{align*} with the convention that $c_{-1}=0$ if the index $-1$ is absent from the expansion. [/step] [step:Pull back the Laurent expansion and isolate the residue term] On a sufficiently small punctured neighbourhood of $b$, substituting $z=\varphi(w)$ gives \begin{align*} f(\varphi(w))\varphi'(w)=\sum_{m=-N}^{\infty} c_m t(w)^m t'(w). \end{align*} The term with $m=-1$ equals \begin{align*} c_{-1}\frac{t'(w)}{t(w)}. \end{align*} Since $t(w)=(w-b)u(w)$ and $u(b)\neq 0$, after shrinking $V_0$ if necessary we have $u(w)\neq 0$ on $V_0$. Therefore \begin{align*} \frac{t'(w)}{t(w)}=\frac{1}{w-b}+\frac{u'(w)}{u(w)}. \end{align*} The function $u'/u:V_0\to\mathbb{C}$ is holomorphic, so the residue of $c_{-1}t'(w)/t(w)\,dw$ at $w=b$ is $c_{-1}$. [guided] The purpose of introducing $t(w)=\varphi(w)-a$ is to measure the coordinate change relative to the point where the residue is taken. Because $\varphi(b)=a$, the function $t:V_0\to\mathbb{C}$ vanishes at $b$. Because $t'(b)=\varphi'(b)\neq 0$, this zero is simple, so we can factor \begin{align*} t(w)=(w-b)u(w), \end{align*} where $u:V_0\to\mathbb{C}$ is holomorphic and $u(b)\neq 0$. Now substitute $z=\varphi(w)$ in the Laurent expansion of $f$ at $a$. Since $z-a=t(w)$ and $dz=\varphi'(w)\,dw=t'(w)\,dw$, the pulled-back differential has coefficient function \begin{align*} f(\varphi(w))\varphi'(w)=\sum_{m=-N}^{\infty} c_m t(w)^m t'(w) \end{align*} on a sufficiently small punctured neighbourhood of $b$. The residue is the coefficient of $(w-b)^{-1}$. The potentially dangerous term is the Laurent term with exponent $m=-1$, because it gives \begin{align*} c_{-1}t(w)^{-1}t'(w)=c_{-1}\frac{t'(w)}{t(w)}. \end{align*} Using the factorisation $t(w)=(w-b)u(w)$, we compute \begin{align*} t'(w)=u(w)+(w-b)u'(w). \end{align*} Dividing by $t(w)=(w-b)u(w)$ gives \begin{align*} \frac{t'(w)}{t(w)}=\frac{1}{w-b}+\frac{u'(w)}{u(w)}. \end{align*} Because $u(b)\neq 0$, the function $u$ is nonzero on some neighbourhood of $b$, and therefore $u'/u$ is holomorphic there. Thus the coefficient of $(w-b)^{-1}$ in $c_{-1}t'(w)/t(w)$ is exactly $c_{-1}$. [/guided] [/step] [step:Show all remaining Laurent terms have zero residue] For every integer $m\geq 0$, the function $t^m t':V_0\to\mathbb{C}$ is holomorphic, so the differential $t(w)^m t'(w)\,dw$ has zero residue at $b$. For every integer $m\leq -2$, define the [meromorphic function](/page/Meromorphic%20Function) \begin{align*} G_m: V_0 \setminus \{b\} \to \mathbb{C}, \qquad w \mapsto \frac{t(w)^{m+1}}{m+1}. \end{align*} Then $G_m$ is meromorphic at $b$, and differentiating gives \begin{align*} G_m'(w)=t(w)^m t'(w). \end{align*} The derivative of a meromorphic function has zero residue: if $G_m(w)=\sum_{k=-M}^{\infty} d_k(w-b)^k$ is its Laurent expansion at $b$, then \begin{align*} G_m'(w)=\sum_{k=-M}^{\infty} k d_k(w-b)^{k-1}, \end{align*} and the coefficient of $(w-b)^{-1}$ would require $k=0$, whose coefficient is $0\cdot d_0=0$. Hence each differential $t(w)^m t'(w)\,dw$ with $m\leq -2$ has zero residue at $b$. [/step] [step:Conclude that the pulled-back residue equals the original residue] The Laurent expansion of $f$ has only finitely many negative-power terms. Among those terms, the preceding step shows that every exponent $m\neq -1$ contributes zero residue after pullback, and the exponent $m=-1$ contributes $c_{-1}$. The holomorphic tail with $m\geq 0$ also contributes zero residue. Therefore \begin{align*} \operatorname{Res}_{w=b}\left(f(\varphi(w))\varphi'(w)\,dw\right)=c_{-1}. \end{align*} Since $c_{-1}=\operatorname{Res}_{z=a}(f(z)\,dz)$, we obtain \begin{align*} \operatorname{Res}_{w=b}\left(f(\varphi(w))\varphi'(w)\,dw\right)=\operatorname{Res}_{z=a}\left(f(z)\,dz\right). \end{align*} This proves the coordinate invariance of the residue. [/step]