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# Proof with Maximum Modulus Principle

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Expand the given sketch of a proof into a reasonably complete proof.
It is suggested that the following proof be used:

Maximum Modulus Principle: Let U be a non-empty open subset of the complex numbers C, let D(z0,r)={z є C: |z-z0| ≤ r} be a closed disk (with center z0 and radius r >0) that is entirely contained in U, let C(z0,r) = {z є C: |z0-z|=r} be the boundary D(zo,r) (a circle), and let f: : U C be an analytic function. The maximum value that |f| attains on D(z0,r) is in fact attained at a point of C(z0,r). Furthermore, if |f| has a non-local maximum on C(z0,r) (equivalently, if |f| attains its maximum value on C(z0,r) at infinitely many points of C(z0,r)), then f is constant on D(z0,r) - so if U is connected, then f is constant on U.

1. The Fundamental Theorem of Algebra states that , if p is a non-constant polynomial with complex coefficients, then p has at least one root: that is, there is at east on complex number z that is a solution of the equation p(z)=0.

Sketch of a proof: suppose p is a polynomial with complex coefficients, and for all complex numbers z, the value p(z) is non-zero. Then the reciprocal 1/p is an analytic function defined on all C. Find upper and lower bounds (depending on r > 0) for the absolute value |1/p| on the circle C(0,r), and then apply the Maximum Modulus Principle to conclude that 1/p-and thus also p- is constant.

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Proof with the maximum modulus principle is modeled.

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Proof:
Suppose for all , . Then we consider . Since is a polynomial with complex coefficient, then is analytic over and thus is also analytic over . From the Maximum ...

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