Fundamental theorem of algebra (proof)
General · Mathematics
Study notes
Q: Prove z²+1 has roots using Liouville. Factor it. Suppose z²+1 ≠ 0 everywhere. Then 1/(z²+1) is ENTIRE. |z| → ∞: 1/(z²+1) → 0: BOUNDED! Liouville: constant - but 1/(0²+1) = 1, 1/(1+1) = 1/2: not constant! Contradiction: z²+1 HAS a root (z = ±i!). Factor: (z-i)(z+i)!