Let P be a polynomial in R^2 and $\oint$ P(x,y)(dx + idy) = 0 for every circle $\gamma$ in $\mathbb{C}$. Prove that P is $\mathbb{C}$-differentiable. I approached this question by first proving the Cauchy integral formula and then using that to get the taylor series expansion. But i am facing difficulty in getting the taylor series expansion.
 A: Here are two ways to make the transition from line integrals to (partial) derivatives:
(I) You can write your polynomial $P$ in the form $P(x,y)=u(x,y)+iv(x,y)$, where $u$ and $v$ are real polynomials in the real variables $x$ and $y$. By assumption
$$\int_\gamma P(x,y)(dx+i\>dy)=\int_\gamma (u+iv)\>(dx+i\>dy)=\int_\gamma (u\>dx-v\>dy)+i\int_\gamma(v\>dx+u\>dy)=0$$
for all circles $\gamma$ in the plane. Consider now a (small) circular disk $D$. Then, by Green's theorem, you have
$$\int_D (u_x-v_y){\rm d}(x,y)=\int_{\partial D}(v\>dx+u\>dy)=0$$
and similarly
$$\int_D (v_x+u_y){\rm d}(x,y)=-\int_{\partial D}(u\>dx-v\>dy)=0\ .$$
Since $D$ is arbitrary and the partial derivatives $u_x$, $\ldots$, $v_y$ are continuous you can draw conclusions. I shall leave that to you.
(II) If you want to avoid Green's theorem you can argue as follows: The polynomial $P$ is its own Taylor expansion at the origin. Therefore we may write
$$P(x,y)=a+bz+c\bar z+o(r)\qquad(r\to0)\ ,\tag{1}$$
where $a$, $b$, $c\in{\mathbb C}$ and we have written the linear terms $px+qy$ using $z=x+iy$ and $\bar z=x-iy$, which is no restriction. Now let $$\gamma_r:\quad t\mapsto re^{it}\qquad(0\leq t\leq2\pi)$$ be the circle of radius $r\ll1$ around $0$. Then
$$\int_{\gamma_r} a\>dz=0,\quad \int_{\gamma_r}bz\>dz=0,\quad \int_{\gamma_r}c\bar z\>dz=c\int_0^{2\pi}re^{-i\phi}\>i r e^{i\phi}d\phi=2\pi i c r^2\ .$$
Using $(1)$ and the main assumption on $P$ it follows that
$$0=\int_{\gamma_r}P(x,y)\>dz=r^2\bigl(2\pi i c+o(1)\bigr)\qquad(r\to0)\tag{2}$$
(the $o$-term picks up another factor $r$ from the length of $\gamma_r$). If $c\ne0$ the right side of $(2)$  is $\ne0$ for all small $r>0$. It follows that $c=0$. But this implies that $P$ is complex differentiable at the origin with $P'(0)=b$.
It is easy to transport this result to any point $z_0\in{\mathbb C}$, and then we are done.
