Suppose $f(z)=\sum_{n\geq0}a_nz^n$ that converges on an open ball centered at the origin with radius $R>0$. Let $|z_0|<R$, then $z=z_0+(z-z_0)$. Apply the binomial expansion on $z^n$, we have
$$z^n=(z_0+(z-z_0))^n=\sum_{k=0}^n\binom{n}{k}z_0^{n-k}(z-z_0)^{k}$$
Thus,
$$f(z)=\sum_{n\geq0}a_nz^n=\sum_{n\geq0}a_n(z_0+(z-z_0))^n=\sum_{n\geq0}a_n\sum_{k=0}^{n}\binom{n}{k}z_0^{n-k}(z-z_0)^{k}$$

I try to simplify the following singular integral:

>$$\int_{R^2}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi= A\int_{R^2}\int_{R^2}p(x) q(y) (-\log|x-y|)dxdy$$
where $p, q\in C_c^{\infty}$ and $\int q=0$.

My way is LHS=
$$\lim_{\epsilon\to 0}\int_{R^2\setminus B(\epsilon)}\hat{p}(\xi)\hat{q}(\xi)|\xi|^{-2}d\xi$$
But why the RHS appears $(-\log|x-y|)$ which is the log potential of Poisson equation.