Relaxing the identity theorem Suppose I have a polynomial in both $z$ and $\overline{z}$ that vanishes (identically) on an open subset of the complex plane. The identity theorem will not apply to such a function since the polynomial is not holomorphic (due to the presence of the $\overline{z}$). 
How would I show that the coefficients of the polynomial must necessarily be zero? 
If we attempt to expand such an expression in terms of $x+iy$, the binomial expansions become horrendous and difficult. 
 A: The fact that our function $f(z, \bar z)$ is a polynomial makes our lives simple.
Assume that the origin is contained inside the open set $U$ on which $f$ is known to vanish. (If not, just redefine the coordinates by a shift.) Let $(x,y)$ be the real coordinates on the complex plane, so $z = x + iy$ and $\bar z = x - iy$. As your function $f(z, \bar z)$ can be expressed as a polynomial in $z$ and $ \bar z$, it certainly can be expressed as a polynomial in $x$ and $y$. It's not necessary to explicitly compute this polynomial like you were suggesting - we only need to know that such a polynomial expression exists. So to show that $f$ is zero everywhere, it suffices to show that the coefficients of $f$ (written as a polynomial in terms of $x$ and $y$) are all zero.
But these coefficients are merely the (higher, mixed) partial derivatives of $f$ with respect to $x$ and $y$, evaluated at the origin. Since $f$ is known to be zero on the open neighbourhood $U$ of the origin, all of these partial derivatives vanish. Therefore, the coefficients of $f$ in terms of $x$ and $y$ must all vanish, and so $f$ is the zero polynomial, and is zero everywhere.
