Show that there exists no sequence of complex polynomials which converges uniformly to $\frac{1}{z^2}$ on the annulus $1<|z|<2$.

Question

Show that there exists no sequence of complex polynomials which converges uniformly to $\frac{1}{z^2}$ on the annulus $1<|z|<2$.

Suppose there exists a sequence of complex polynomials $f_n(z)=a_0^{(n)}+a_1^{(n)} z+a_2^{(n)} z^2+\cdots +a_n^{(n)}z^n$ which converge uniformly to $\frac{1}{z^2}$ where $a_n^{(n)}\neq 0$

Then $\sup_{1<|z|<2}|f_n(z)-\frac{1}{z^2}|<\epsilon $ $\forall \epsilon>0\to (\ast)$.

Now note that at $1<|z|<2 ,|z|^n\to \infty$

Thus $|f_n(z)|\to \infty $ as $n\to \infty $ making $f_n(z) $ unbounded but $\frac{1}{z^2}$ is bounded in $1<|z|<2$.

Thus $(\ast) $ cant hold $\forall \epsilon>0$.

Thus no such sequence of polynomial exists.

Answer 1

The OP's argument has a gap in that the order of growth of $|a_n^{(n)}z^n + a_{n-1}^{(n)}z^{n-1}+a_0^{(n)}|$ not only depends on $z^n$, but also on the magnitudes of the leading coefficient $a_n^{(n)}$ and terms of the lower degrees. The $j$-th coefficients $a_j^{(n)}$ varies as $n$ varies and we have little information on their magnitudes. This precludes us from drawing the conclusion that $|f_n(z)|\to\infty$ as $n\to\infty$.

Hint: Note that $$ \frac{1}{2\pi i}\int_{|z|=1.5}\frac{1}{z^2}\cdot zdz=1 $$ while $$ \frac{1}{2\pi i}\int_{|z|=1.5}p(z)\cdot zdz =0 $$ for all polynomial $p(z)$. Show that if there is a sequence of polynomials $p_n(z)$ uniformly converging to $\frac{1}{z^2}$ on $1<|z|<2$, then it holds$$ \frac{1}{2\pi i}\int_{|z|=1.5}\frac{1}{z^2}\cdot zdz=0, $$ which leads to a contradiction.

Answer 2

I am not sure if I am correct, but please correct me and let me know my mistake if possible, probably there is a huge thing I am missing.
My solution is like, see that $z^{-2}$ attains $\frac{1}{4}$ on the outer boundary of the annulus $(|z| = 2)$ and it attains 1 at inner boundary $(|z| = 1)$. Then we know polynomials are entire. Then if there exists such a sequence, it is entire and uniformly is approximating $z^{-2}$. Then it will be close to the same values as $z^{-2}$ over boundaries, but this is not possible by maximum modulus principle(cause entire function in $|z| < 2$ must attain its maximum on boundary $|z| = 2$, unless it is a constant function. But we see polynomials have larger value at $|z| = 1$, so they must be constant, which is contradiction.