1

If I have a polynomial of the form $p(z) = a_0 + a_1z + \cdots + a_nz^n$ with complex coefficients, how can I show that each coefficient is bounded.

M. Smith
  • 137

1 Answers1

2

$$a_k = \text{Res}_{x=0}\frac{p(x)}{x^{k+1}} = \frac{1}{2\pi i}\oint_{|z|=1}\frac{p(z)}{z^{k+1}}\,dz\tag{1} $$ by Cauchy integral formula / the residue theorem. By switching to absolute values: $$ |a_k|\leq \frac{1}{2\pi}\oint_{|z|=1}\frac{|p(z)|}{|z|^{k+1}}\,dz \leq \frac{1}{2\pi}\oint_{|z|=1}|p(z)|\,dz \leq \max_{|z|=1}|p(z)|\tag{2}$$ as wanted.

Jack D'Aurizio
  • 353,855
  • I'm unfamiliar with the residue theorem, is there a way of solving the problem without using that? I think perhaps you could come to the same conclusion using Cauchy's estimate? – M. Smith Mar 28 '17 at 02:08
  • @M.Smith:just remove the $=\text{Res}_{x=0}\ldots=$ part from $(1)$ and the argument still applies. – Jack D'Aurizio Mar 28 '17 at 02:14