Suppose that $f$ is analytic in the annulus: $1 \leq \vert z \vert \leq 2 $, that $\vert f \vert \leq 1$ for $\vert z \vert = 1$ and that $\vert f \vert \leq 4$ for $\vert z \vert = 2$. Prove $\vert f(z) \vert \leq \vert z \vert ^2$ throughout the annulus.
I know that I would have to apply the Maximum Modulus Theorem here, but I am having trouble figuring out how to do so. Would I have to use the analyticity of $f$ in order to reach such a conclusion?
I am using the textbook Complex Analysis, Third Edition by Joseph Bak and Donald J. Newman.
Any suggestions and tips are greatly welcomed.
As suggested, my comment above will be posted as an answer to seek clarification. If $z=1$, then it the inequality holds. If $z=2$, the inequality is still true. The Maximum Modulus Theorem states a non-constant analytic function in a region $D$ does not have any interior maximum points. Using $\frac{f(z)}{z^2}$, then it does not have any interior maximum points; hence, it assumes its maximum modulus at its boundary points?