So I have a question regarding the Wirtinger derivative with regards to the polynomial $p(z)=2z^2\bar{z}^3-9z\bar{z}^2+z^2+12\bar{z}+1$. The original problem is to determine for which $z_0 \in \mathbb{C}$ such that $p'(z_0)$ exists. By Goursat's theorem, this is equivalent to finding a domain $D = \{z \in \mathbb{C} : p(z) \ \text{analytic}\}$.
$p(z)$ is analytic whenever the Cauchy-Riemann equations are satisfied, which, using the Wirtinger derivatives corresponds to solving the equation $\frac{\partial p}{\partial \bar{z}} = 0$.
The differential operator $\frac{\partial}{\partial \bar{z}}$ is defined as $\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left[\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right]$ with $z = x+iy$ and $\bar{z} = x-iy$.
Is it permitted to differentiate $p(z)$ with respect to $\bar{z}$ using the standard differentiation rules for polynomials, i.e. obtaining $\frac{\partial p}{\partial \bar{z}} = 6z^2\bar{z}^2 - 18z\bar{z} + 12$? Or do I have to rewrite $p(z)$ in terms of $x$, and $y$, in order to apply the Wirtinger differential operator?
A way of checking this is to compare directy, i.e. $0 = \frac{\partial p}{\partial \bar{z}} = 6z^2\bar{z}^2 - 18z\bar{z} + 12 = 6|z|^4 - 18|z|^2 + 12 \iff |z| = 1, \text{or}\ |z|=\sqrt{2}.$$
And by rewriting $p(z)$ as $$p(x,y) = 2|z|^4(x-iy) - 9|z|^2(x-iy) +z^2 + 12\bar{z} + 1 = 2(x^5 - iyx^4 + y^2x^3 - 2iy^3x^2 + y^4x - iy^5) - 9(x^3-iyx^2+xy^2-iy^3) + (x^2 - y^2 + 2xyi) + 12(x-iy) + 1,$$ but in this case the calculations get messy quickly. All feedback is welcomed.
Thanks!
$\frac{\partial}{\partial\bar z}$ satisfies Leibniz rule. It follows directly from linearity and $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ satisfying Leibniz rule.
Also, for any holomorphic function $f$, as you said, we have $\frac{\partial}{\partial\bar z}f(z) = 0$. So, if you have polynomial $\sum_n a_n(z)\bar z^n$ in $\bar z$ with coefficients holomorphic functions $a_n(z)$, then
$$\frac{\partial}{\partial\bar z}\sum_n a_n(z)\bar z^n = \sum_n \left(\frac{\partial}{\partial\bar z}(a_n(z))\bar z^n+a_n(z)\frac{\partial}{\partial\bar z}(\bar z^n)\right) = \sum_nna_n(z)\bar z^{n-1}.$$
To say it simply, holomorphic functions are like constants when you are acting by $\frac{\partial}{\partial\bar z}$. So, yeah, you can differentiate your function just as a polynomial in $\bar z$ with $z$ being a constant.
Writing your function in $\{x,y\}$ coordinates will give you the same result:
$$p(x,y) = 1 + 12 x + x^2 - 9 x^3 + 2 x^5 - y^2 - 9 x y^2 + 4 x^3 y^2 + 2 x y^4 + i(-12 y + 2 x y + 9 x^2 y - 2 x^4 y + 9 y^3 - 4 x^2 y^3 - 2 y^5)$$ and $$\frac 12\left(\frac{\partial}{\partial x}+ i \frac{\partial}{\partial y}\right)p(x,y) = 6(x^2+y^2)^2-18(x^2+y^2)+12.$$