Maximum value estimation

Question

Let $f$ be an analytic function that is not zero at $\{z:|z|<2\}$.

Show that for every natural number $n$: $$\max_{|z|=1}|f(z)-\frac{1}{z^n}|>1$$

I know that $|f(z)-\frac{1}{z^n}|=|f(z)z^n-1|$ on the unit circle and I think I need to use the maximum modulus principle but I don't see how to proceed. I would like a hint

Answer 1

Since the inequality need not hold for $n = 0$, I assume that $0$ doesn't count as a natural number here.

Then note that $z^nf(z)$ has a zero (of order $n > 0$) in $0$, so

$$\sup_{\lvert z\rvert < 1} \lvert z^n f(z)-1\rvert \geqslant 1.$$

Now use the maximum modulus principle.

Answer 2

In a neighborhoood of zero one has $f(z)z^n=f(0)z^n+O(|z|^{n+1})=ce^{in\phi+\phi_0}r^n+O(r^{n+1})$, where $r=|z|$, $\phi=\operatorname{arg}z$, and $f(0)=ce^{i\phi_0}$, $c>0$ by the condition imposed on $f$. Take $z=re^{i(\pi-\phi_0)/n}$, then for small $r$ one has $|z^nf(z)-1|>1$, and it remains to use the maximum modulus principle.