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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.