Fundamental Theorem of Algebra

Complex Analysis · Axiom Academy

LESSON Fundamental Theorem of Algebra Why every non-constant polynomial must have a root — and how complex analysis proves it in one stroke. 1. The Claim: Roots Always Exist in Take any polynomial with complex coefficients and degree . The theorem promises a point z_0 in the complex plane where p(z_0) = 0 . Counting multiplicity, it actually has exactly n roots, so it factors completely into linear pieces. A degree- n polynomial splits into n linear factors over Watch it happen for p(z) = z^3 - 1 . Its three roots are the cube roots of unity — they sit on the unit circle, evenly spaced apart: 1 on the real axis, and at . 2. The Key Tool: Liouville's Theorem Here is the single fact the proof rests on. A function is entire if it is complex-differentiable on all of . Liouville's theorem says such a function has nowhere to hide: This is shocking from a real-variable point of view — is smooth and bounded yet far from constant. Complex differentiability is far more rigid: being analytic everywhere and staying inside a finite band leaves no room to wiggle. 3. The Contradiction: 1/p Has Nowhere to Go Suppose, for contradiction, that for every z . Then g(z) = 1/p(z) is defined everywhere and is entire . We just need to show it is bounded — then Liouville finishes the job.

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