I am looking at number 1b of this released exam:

Suppose $f(z) = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4$. Show that if: $$ |f(w^j)| \leq A; \space\space j=0,1,2,3,4$$ where $w = e^{2\pi i/5}$ then $|a_0|\leq A$.

I'm pretty sure we're supposed to use part (a). So:

$$f(z) = \sum_{k=0}^4 a_kz^k \mapsto \sum_{k=0}^4 a_k(w^j)^k$$

Then we can say that:

$$\sum_{k=0}^4 a_k(w^j)^k \leq \max{a_k} \sum_{k=0}^4 (w^j)^k $$

But from (a), $\sum_{k=0}^4 (w^j)^k = 0$ so:

$$\sum_{k=0}^4 a_k(w^j)^k \leq 0$$

I don't know where to go from here though to conclude the desired result. Any help would be appreciated.


1 Answer 1


\begin{align*} \sum_{k=0}^4 f(\omega^k) &= 5a_0 + a_1\sum_{j=0}^4 \omega^j + a_2 \sum_{j=0}^4 \omega^{2j} + \cdots + a_4\sum_{j=0}^4 \omega^{4j} \end{align*} From $(a)$ of the problem, all terms except $5a_0$ vanishes. Thus \begin{align*} |5a_0| &= \left| \sum_0^4 f(\omega^k) \right| \leq\sum_0^4 |f(\omega^k)| \leq 5A \end{align*} and hence $|a_0| \leq A$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .