Approximating $1/z$ by polynomials Let $C=\{\mathrm e^{\mathrm it}, 0\le t\le 3\pi/2\}$ and $f(z)=1/z$. By Runge's theorem, there is a sequence of polynomials $p_n(z)$ such that $$\lim_n p_n(z)=f(z)$$ uniformly on $C$. 
Does anyone know such a sequence?
 A: This solution is specialized to the particular problem. As in my other solution, I am working with the arc $C = \{ e^{i \theta} : \pi/4 < \theta < 7 \pi/4 \}$. Let $T_n(z)$ be the $n$-th Chebyshev polynomial, so it is the polynomial with leading term $2^{n-1} z^n$ which has $|T_n(z)|\leq 1$ for $-1 \leq z \leq 1$. 
Set
$$g_n(z) = z^{-1} - \frac{(1 + 1/\sqrt{2})^n  z^{n - 1}}{2^{n-1}}
 T_n\left(\frac{z + z^{-1} +1-1/\sqrt{2}}{1 + 1/\sqrt{2}} \right)$$
Okay, what's going on here? The map $z \mapsto z + z^{-1}$ takes $C$ to $[-2, \sqrt{2}]$. The linear function inside the Chebyshev polynomial takes $[-2, \sqrt{2}]$ to $[-1,1]$, so the $T_n$ term is $\leq 1$. So the whole second term has absolute value at most $(1+1/\sqrt{2})^n/2^{n-1} \approx 2 \cdot 0.853^n$. This will be much less than $z^{-1}$, for $n$ large, so $g_n(z) \approx z^{-1}$. 
On the other hand, $T_n\left( \mbox{stuff} \right)$ will be a Laurent polynomial with most negative term $\frac{2^{n-1} z^{-n} }{(1+1/\sqrt{2})^n}$. 
After multiplying by $\frac{(1 + 1/\sqrt{2})^n  z^{n - 1}}{2^{n-1}}$, the second term will be of the form $z^{-1} + \mbox{polynomial}$. So $g_n(z)$ is a polynomial. 

In this picture, the black arc is $e^{i \theta}$ for $\pi/4 \leq \theta \leq \pi$. The red, blue and green arcs are $g_{10}$, $g_{20}$ and $g_{30}$ evaluated on the black arc. 
How did I come up with this? Roughly: I want $z^{-1} \approx p(z)$ for a polynomial $p$. I want $z^{-1} - p(z) \approx 0$. I want $z^{-n} + \cdots + z^n \approx 0$. I want Laurent polynomials with leading term $1$ and very small values on $C$. I know a family of polynomials with leading term $1$ and very small values on $[-1, 1]$: Namely, $2^{-n+1} T_n(z)$. How can I turn one into the other?
Note that these have much smaller degree than the polynomials in my other solution. The thing I called $f_N$ in my other solution has degree $N^3$, this $g_n$ has degree $2n-1$.
A: I'm going to put up two answers to this problem. I find it more convenient to rotate the curve $C$ to be $\{ e^{i \theta} : \pi/4 \leq \theta \leq 7 \pi /4 \}$. The first one is the standard proof of Runge's theorem, made concrete.
Notice that
$$\frac{1}{z} = - \sum_{k=0}^{\infty} 0.6^k  \left( \frac{1}{0.6 - z} \right)^{k+1}$$
$$\frac{1}{0.6-z} = \sum_{\ell=0}^{\infty} 0.6^{\ell} \left( \frac{1}{1.2-z} \right)^{\ell}$$
$$\frac{1}{1.2-z} = \sum_{m=0}^{\infty} (1.2)^{-m-1} z^m$$
are all convergent on $C$. The key point is that $0.6$ is closer to $0$ than it is to any point on $C$; $1.2$ is closer to $0.6$ than to any point on $C$ and $1.2$ is outside the circle $C$. Plugging the third equation into the second into the first gives
$$\frac{1}{z} = - \sum_{k=0}^{\infty} 0.6^k \left( \sum_{\ell=0}^{\infty} 0.6^{\ell} \left( \sum_{m=0}^{\infty} 1.2^{-m-1} z^m \right)^{\ell+1} \right)^{k+1}$$
Define 
$$f_N(z) = - \sum_{k=0}^{N} 0.6^k \left( \sum_{\ell=0}^{N} 0.6^{\ell} \left( \sum_{m=0}^{N} 1.2^{-m-1} z^m \right)^{\ell+1} \right)^{k+1}$$
Then $f_N(z) \to z^{-1}$ on $C$.

In the image, the red arc is $z$ for $\pi/4 \leq \theta \leq \pi$. The green arc is $f_{20}(z)$ for $z$ in the same range; the blue arc is $f_{50}(z)$. As you can see, the blue arc is nearly indistinguishable from a true circle.
This method should be clearly adaptable to take any function on $C$ with a pole at $0$ and write it as a limit of polynomials. For other curves and other pole locations, you'd replace $(0.6, 1.2)$ with a different sequence of complex numbers, but the idea remains the same.
A: Let me show how to first construct a sequence of polynomials $p_n$ such that $p_n(z)\to1/z$ uniformly on $C=\{e^{it}\colon \theta\le t\le 2\pi-\theta\}$, for any given $\theta\in(0,\pi)$.
Choosing real numbers $0 < a < b < 1$, set
$$
u(z)=\frac{1-az}{1-bz}.
$$
For $z=e^{i\phi}$ this gives
$$
1-\lVert u(z)\rVert^2 = \frac{(b-a)\left(a+b-2\cos\phi\right)}{1-2b\cos\phi+b^2}.
$$
So, $\lVert u\rVert < 1$ on $C$ so long as $(a+b)/2 > \cos\theta$. Now expand out as a power series and truncate after some number $m$ terms.
$$
\begin{align}
v(z)&=(1-az)\left(1+bz+b^2z^2+\cdots+b^mz^m\right)\\
&=u(z)(1-b^{m+1}z^{m+1})
\end{align}.
$$
On $C$, this is bounded by $\lVert u(z)\rVert(1+b^{m+1})$ so, for large enough $m$, $\lVert v\rVert < 1$ on $C$. As $v(0)=1$, the following is a sequence of polynomials,
$$
p_n(z)=z^{-1}\left(1-v(z)^n\right).
$$
As $\lVert v\rVert < 1$ we have $p_n(z)\to z^{-1}$ uniformly on $C$. To deal with the range $\{e^{it}\colon0\le t\le3\pi/2\}$, we can take $\theta=\pi/4$ above and rotate the set $C$ by an angle of $\pi/4$. That is,
$$
p_n(z)=z^{-1}\left(1-v(e^{\pi i/4}z)^n\right).
$$
To state some concrete numbers, taking $a=1/\sqrt2$, $b=0.78$ and $m=21$ works, so that $\lVert v\rVert < 0.999$ on $C$.
