show residue of a holomorphic function at a singularity is equal to composite of a locally bilholomorphic function times its derivative Suppose $\phi$ maps a neighborhood of $p$ biholomorphically to a neighborhood of $q=\phi(p)$. Show for every holomorphic function $f$ with isolated singularity $q$, we have $$res(f,q)=res((f\circ\phi)\phi^\prime,q)$$
Suppose $W,V$ are neighborhood of $p,q$ respectively, let $D_r(q)\subset V$ then we have $res((f\circ\phi)\phi^\prime,q)=\frac{1}{2\pi i}\int_{\mathbb{T}_r(q)}(f\circ\phi)(z)\phi^\prime(z) dz$  and $res(f,p)=\frac{1}{2\pi i}\int_{\mathbb{T}_s(p)}f(z)dz$ where $D_s(p) \subset W$. Denote $w=\phi(z)$ then $\phi^\prime(z)=wdw$. WTS that
$\int_{\mathbb{T}_r(q)}f(w)w dw=\int_{\mathbb{T}_s(p)}f(z)dz$.
We have $\int_{\mathbb{T}_s(p)}f(z)dz=\int^{2\pi}_0f(p+se^{it})ise^{it}dt$, I am stuck here.
Can someone help me out here with some hints or give me an outline of what should I try to show?

Start from the integral you wrote down for the residue, and try the change of variables $w = \phi(z)$.
You should also write down a similar integral that defines $res(f, p)$. You'll be done if you can show those two integrals are equal.
