What are the maximum moduli of (a) $f(z) = \cos3z$ and (b) $g(z) = \frac{2z+1}{2z-1}$ on $\bar B(0,1)$?
Attempt:
Clearly we will want to use the maximum modulus principle for (a) and for (b) I am not sure at all, as $g$ has a simple pole at $z= \frac{1}{2}$, which is inside $\bar B(0,1)$.
If anyone could help me out here that would be great! Many thanks!
As you have observed $g$ is not defined at $\frac 1 2$ so b) does not make sense.
For a) write $z$ on the boundary as $e^{it}$ with $t$ real. You can see that $|cos 3z| \leq \sup_t \frac {e^{-3\sin t} +e^{3\sin t}} 2$. Now use the fact that $\frac {x+\frac 1 x} 2$ is increasing in $[-1,1]$ to see that the maximum is attained when $\sin t=1$. Hence $|\cos 3z| \leq \frac {e^{-3}+e^{3}} 2$ . This value is attained when $t=\frac {\pi} 2$ which means $z=i$. Hence the answer to a) is $\frac {{e^{-3}+e^{3}}} 2$ .