Proving that the space of polynomials in $z$ on the unit ball $B(0,1) \subset \mathbb{C}$ is not dense in $C(B(0,1)) $ Without using complex analysis, is it possible to prove that the set of 
polynomials in $z$ is not dense in $ C(B(0,1)) $ (with the uniform norm)? 
Using Morera's theorem, the closure is just the set of analytic functions, but is there an easier way to see this? For instance, can one prove that $ z \mapsto \overline{z} $ is not in the closure, just by using techniques of real analysis? 
 A: Hint: Let $f(z) = \bar z -\sum_{k=0}^{n}c_kz^k.$ What is
$$ \int_0^{2\pi}|f(e^{it})|^2\ dt?$$
A: In the spirit of zhw's answer:
Take $f(z) = |z-1|^2$, and let $\gamma(t) = r e ^{i t}$.
A tedious integration gives $\int_\gamma f(z) dz = -2 \pi i r$.
$| \int_\gamma (f(z) - p(z)) dz | =2 \pi r \le \int_\gamma |f(z) - p(z)| |dz| \le 2 \pi \|f-p\|_\infty$.
In particular, this gives $\|f-p\|_\infty \ge 1$.
Note: In some sense this is Morera's theorem. Given any suitable $f$ that is not analytic, we can find some curve $\gamma$ such that $\int_\gamma f(z)dz \neq 0$, from which a result similar to the above follows. The above has the nice property that it gives an explicit lower bound to the approximation.
A: Yet another similar approach, secretly using Cauchy's integral theorem.  I suppose here that $B(0,1)$ means the closed unit ball (otherwise make obvious modifications).
Consider the functional $I : C(B(0,1)) \to \mathbb{C}$ defined by 
$$I(f) = \int_0^1 f(e^{2 \pi i t}) e^{2 \pi i t}\,dt$$
It's clearly a linear functional, and $|I(f)| \le \|f\|_\infty$ so it's continuous.  It's also not the zero functional by noting $I(\bar{z})=1$.    Now verify that $I(z^n) = 0$ for every $n$.  By linearity, $I(f)=0$ for every polynomial in $z$.  So all the polynomials in $z$ are contained in the kernel of $I$, which, since $I$ is continuous, is a proper closed linear subspace of $C(B(0,1))$.  Moreover, since $\bar{z}$ isn't in the kernel of $I$, it's not in the closure of the polynomials in $z$.
A: Any polynomial must have a leading term, which will be $cz^n$ for some complex $c$ and some non-negative integer $n$. For certain chosen angles and large magnitudes of $z$, you can show that this does not approach $f(z) = f(a + bi) = a - bi$ in the ratio limit, so they cannot be equal.
A: You can look at the integral over the curve $\gamma = \{z: |z| = {1 \over 2}\}$. Directly parameterizing shows that the integral over any $z^n$ $(n \geq 0)$ over $\gamma$ is zero. Hence the same is true for any polynomial. On the other hand, the integral of $\bar{z} = {1 \over 4 z}$ over $\gamma$ is ${\pi i \over 2}$. 
But the integral of the limit is the limit of the integral under uniform convergence. Each polynomial integrates to zero over $\gamma$, so the same must be true for any uniform limit of polynomials. Hence $\bar{z}$ is not the uniform limit of polynomials.
