I've been working on the following exercise:
Can the function $\text{Re}(z^2)$ be approximated uniformly on the unit circle $\{z\,:\,|z|=1\}$ be rational functions having only simple poles? Explain why or why not.
I'm of the belief that this is not true. My intuition is the following:
On the unit circle we have $\bar{z}=1/z$, so
$$
\text{Re}(z^2)=2z^2+2\bar{z}^2=2z^2+\frac{2}{z^2}
$$
on the unit circle. Now the function $z^2+z^{-2}$ has a pole of order two at the origin, so I would think one could get a contradiction by exploiting the residue theorem.
Attempt
More precisely, here is the argument I'd like to say. Suppose such a sequence of rational functions $\{r_n\}_{n=1}^{\infty}$ exists. By looking at the sequence of integrals $\int_{|z|=1}\frac{r'_n}{r_n}dz$, the argument principle implies that the number of poles of $r_n$ within the unit disc is eventually constant. By pole-pushing, we may assume that these singularities are located at the points $a_1,\ldots a_k$, where $a_1=0$, for each $r_n$. Define the Blaschke Factors by
$$
B(z,a)=
\begin{cases}
z, &\text{ if }a=0\\
\frac{z-a}{1-\bar{a}z}, &\text{ if }a\neq 0
\end{cases},
$$
where $a$ is in the unit disk. Then the sequence $\{\tilde{r}_n\}_{n=1}^{\infty}:=\left\{\prod_{j=1}^{k}B(z,a_j)r_n(z)\right\}_{n=1}^{\infty}$ approximates $\left(\prod_{j=1}^{k}B(z,a_j)\right)\left(z^2+\frac{1}{z^2}\right)$ uniformly on the unit circle. However, since $a_1=0$, the function $\left(\prod_{j=1}^{k}B(z,a_j)\right)\left(z^2+\frac{1}{z^2}\right)$ has a simple pole at $z=0$, and so
$$
\int_{|z|=1}\prod_{j=1}^{k}B(z,a_j)\left(z^2+\frac{1}{z^2}\right)dz\neq 0.
$$
However each $\tilde{r}_n$ is actually holomorphic on the unit disk, so that Cauchy's theorem yields a contradiction.
$$
$$
Basically, I'm trying to "jazz up" the proof in the case that each $r_n$ has a simple pole at $z=0$, but this is considerably more difficult than I would've thought.
Error in Attempt
You may or may not have caught it, but there is a bit of a technical complication in my proof when I used the argument principle to show that the number of poles of $r_n$ is eventually constant. The problem comes from the fact that $z^2+1/z^2$ has zeros on the unit circle, so the integral $\int_{|z|=1}r_n'/r_n$ may "blow up" as $n\to\infty$.
Does anyone have any other approaches? Or any ways to get around this complication? Thanks a bunch.
Best Answer
Note that $z=\frac{1}{2\pi i}\int_{|\zeta =2|}\frac {\zeta}{\zeta-z}d\zeta, |z| \le 1$
In particular by dividing $|\zeta|=2$ in small chunks and using uniform continuity, one can find rational functions $R_k(z)=\sum_m \frac{c_{k,m}}{z-w_{k,m}}$ with simple poles $|w_{k,m}|=2$ s.t $R_k(z) \to z$ uniformly on $|z| \le 1$. This means $\bar zR_k(\bar z)\to \bar z^2$ uniformly on $|z| \le 1$
But $\bar zR_k(\bar z)=\sum_m \frac{c_{k,m}\bar z}{\bar z-w_{k,m}}$ so if we restrict to $|z|=1$ we get $\bar zR_k(\bar z)=\sum_m \frac{c_{k,m}}{1-w_{k,m}z}=Q_k(z)$, with $Q_k$ rational function with simple poles on $|z|=1/2$ this time, $Q_k(z) \to \bar z^2$ uniformly on $|z|=1$
Obviously $\frac{z^2+Q_k(z)}{2} \to \Re z^2$ uniformly on $|z|=1$ and is of the required type so we are done!
Note - it is trickier but doable to write $z=\sum_{k \ge 1}\frac{c_k}{z-z_k}, |z| \le 1, \sum |c_k| < \infty, |z_k| >\delta > 1$ with uniform convergence on the closed disc, from which the result follows by taking half of $z^2+\sum_{k \ge 1}\frac{c_k}{1-z_kz}$ as before, so $Q_m$ has same poles and residues as $Q_k$ plus $m-k$ more for all $m>k$