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?

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 $C$.

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 $2n-1$.