# Show $|f(z)|<3$ for each $z\in B(0,1)$?

If $f:B(0,2)\to\Bbb C$ be an analytic function satisfying $|f(z)-2|<1$ for each $z\in \Bbb C$ such that $|z|=1$. show that

(a) $|f(z)|<3$ for each $z\in B(0,1)$?

(b) $f(z)\neq0$ for each $z\in B(0,1)$>

I tried to use the formula $f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-z}dz$ on $z\in B(0,1)$ since $f$ is analytic and $\gamma=e^{it}$ to get the bounds. However, it does not work at all.

I really have no clue about this question. Could someone kindly help? Thanks!

Hint: $|f(z) - 2| < 1$ for $|z|<1$ by the maximum modulus theorem.