Prove the Fundamental Theorem of Algebra (using complex analysis).
Solution
- Strategy: By contradiction with Liouville’s Theorem
- Suppose \(p\) is non-constant and has no roots.
- Claim: \(1/p(z)\) is a bounded
holomorphic function on \(\CC\).
Holomorphic: clear? Since \(p\) has no roots.
Bounded: for \(z\neq 0\), write \[\begin{align*} \frac{P(z)}{z^{n}}=a_{n}+\left(\frac{a_{n-1}}{z}+\cdots+\frac{a_{0}}{z^{n}}\right) .\end{align*}\]
The term in parentheses goes to 0 as \(\abs{z}\to \infty\)
Thus there exists an \(R>0\) such that \[\begin{align*} \abs{z} > R \implies \abs{P(z) \over z^n} \geq c \definedas {\abs{a_n} \over 2} .\end{align*}\]
So \(p\) is bounded below when \(\abs{z} > R\)
Since \(p\) is continuous and has no roots in \(\abs{z} \leq R\), it is bounded below when \(\abs{z} \leq R\).
Thus \(p\) is bounded below on \(\CC\) and thus \(1/p\) is bounded above on \(\CC\).
- By Liouville’s theorem, \(1/p\) is constant and thus \(p\) is constant, a contradiction.