# Show that for $f$ analytic in $B(0,2)$, $\max_{|z|=1}|\frac{1}{z}-f(z)|\ge 1$?

Let $f:B(0,2)\to \Bbb C$ be an analytic function. Show that $$\max_{|z|=1}\left|\frac{1}{z}-f(z)\right|\ge 1.$$

I tried to write $f(z)$ as power series since it is analytic, it doesn't seem work. I also tried to use maximum modulus principle, it also did not work.

Could anyone kindly help? I am really struggling with this problem. Thank you!

Hint: Try integrating $1/z-f(z)$ around the unit circle.
• I get $2\pi i-\int_{\gamma}f(z)dz=2\pi i -i\int_0^{2\pi}e^{it}f(e^{it})dt$, I am really confused what to do next. Could you help give some hint about what to do next? Thanks! – Sherry Jul 22 '16 at 19:12
• Well, $f(z)$ is analytic on the entire unit disk. What does that tell you about $\int_\gamma f(z)dz$? – Eric Wofsey Jul 22 '16 at 19:14