A bounded entire function must be a constant function

1. Fundamental Theorem of Algebra

Theorem 1.1. Let P (z) be a complex polynomial of degree d ≥ 1. Then P (z) has a root.

To prove the theorem, we need the Liouville’s theorem.

Theorem 1.2. A bounded entire function must be a constant function.

Let us assume P (z) has no zero. Define f (z) = 1/P (z). By definition, f (z) is an entire function. Write P (z) = adz^d+ · · · + a1z + a0 with ad6= 0. Then we know

z→∞lim P (z)

z^d = lim

z→∞



a_d+a_d−1

z + · · · + a₀ z^d



= a_d. Choose  = |a_d|/2. Then there is A > 0 so that for all |z| > A,

P (z) z^d − a_d

< |a_d| 2 . Using triangle inequality,

P (z) z^d − a_d

≥ |a_d| −    

P (z) z^d

    . This gives |P (z)| > 1

2|a_d||z|^d, for |z| > A. In other words,

|f (z)| =    

1 P (z)

< 2

|a_d||z|^d < 2

|a_d|A^d, |z| > A.

Since {|z| ≤ A} is a compact subset of C, and f is continuous on {|z| ≤ A}, f attains is maximum on {|z| ≤ A}. Let M⁰ be the maximal of f on {|z| ≤ A} and M⁰⁰ = 2/|a_d|A^d. Take M = max{M⁰, M⁰⁰}. Then

|f (z)| ≤ M, ∀z ∈ C.

This shows that f is a bounded entire function. By Liouville Theorem, f must be a constant function. Hence P (z) is a constant polynomial which leads to a contradiction to the assumption that deg P ≥ 1.

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