Use Rouche’s theorem to prove the Fundamental Theorem of Algebra.
Solution
Write \(f(z) = \sum_{k\leq n} c_k z^k\). Big: \(M(z) = c_nz^n\). Small: \(m(z) = f(z) - M(z) = \sum_{k\leq n-1} c_k z^k\).
Now use that \[ \abs{m(z) \over M(z)} &\da \abs{\inverseof{c_n} \sum_{k\leq n-1} c_k z^{k-n}} \\ &= \abs{\inverseof{c_n}\qty{ {c_1\over z^n} + {c_2\over z^{n-1} } + \cdots + {c_{n-1}\over z} }} \\ &\convergesto{\abs{z}\to\infty}0 ,\] so choose \(R\) large enough such that for \(\abs{z} \geq R\), \(\abs{M(z)\over m(z)} < 1\). Then on \(\abs{z} = R\), \[ \abs{m(z) \over M(z)} < 1 \implies \abs{m(z)} < \abs{M(z)} \implies \size n = \size Z_{M} = \size Z_{M+m} = \size Z_{f} ,\] since \(c_n z^n\) has \(z=0\) as a root with multiplicity \(n\).
Solution: Explicit bound
An estimate: write \(f(z) = \sum_{k\leq n} c_k z^k\) with \(c_n = 1\), then for \(R> 1\), on \(\abs{z} = R\) we have \[ \abs{f(z) - z^n} &\leq \sum_{k\leq n-1} \abs{ c_k z^k} \\ &\leq \sum_{k\leq n-1} \abs{ c_k} R^k \\ &\leq \sum_{k\leq n-1} \abs{ c_k} R^{n-1} \\ &= R^{n-1} \sum_{k\leq n-1} \abs{ c_k} \\ &\da R^{n-1} C \\ &\leq R^n \\ &= \abs{z^n} ,\] provided we can choose \(C<R\), but this is possible since \(\sum_{k\leq n-1}\abs{c_k}\) is a constant. So \(n = \size Z_{z^n} = \size Z_f\).