[複變] Liouville's Theorem & The Fundamental Theorem of Algebra

Theorem 1 :(Liouville's theorem)
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If f is entire and bounded in the complex plane, then f(z) is constant throughout the plane.


Theorem 2 :
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Any polynomial

P(z) = a₀ + a₁z + a₂z² + … + anzⁿ　　(an ≠ 0)
of degree n (n≥1) has at least one zero. That is, there exists at least one point z₀ such that P(z₀) = 0.

pf:
Prove by contradiction.
Suppose that P(z) is not zero for any value of z. Then the reciprocal

f(z) = 1/P(z)
is clearly entire, and it is also bounded in the complex plane.
To show that it is bounded, we first write

w = a₀/zⁿ + a₁/zⁿ⁻¹ + a₂/ zⁿ⁻² + … + an₋₁/z,
so that P(z) = (an + w)zⁿ. We then observe that a sufficiently large positive number R can be found such that the modulus of each of the quotients in w = a₀/zⁿ + a₁/zⁿ⁻¹ + a₂/ zⁿ⁻² + … + an₋₁/z is less then the number |an|/(2n) when |z| ≥ R. The generalized triangle inequality, applied to n complex numbers, thus shows that |w| ≤ |an|/2 for such values of z. Consequently, when |z| ≥ R,

|an + w| ≥ ||an| - |w|| > |an|/2;
and this enables us to write

|P(z)| = |an + w||zⁿ| > (|an|/2)|zⁿ| ≥ (|an|/2)Rⁿ

whenever |z| ≥ R.
Evidently, then,

|f(z)| = 1/|P(z)| < 2/(|an|Rⁿ)

whenever |z| > R.
So f is bounded in the region exterior to the disk |z| ≤ R. But f is continuous in that closed disk, and this means that f is bounded there too. Hence f is bounded in the entire plane. It now follows form Liouville’s theorem that f(z), and consequently P(z), is constant. But P(z) is not constant, and we have reached a contradiction.