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单词 ProofOfFundamentalTheoremOfAlgebraargumentPrinciple
释义

proof of fundamental theorem of algebra (argument principle)


The fundamental theorem of algebraMathworldPlanetmath can be proven using the argument principle. Not only is this proof interesting because it demonstrates an important result, it also serves as an example of how to use the argument principle. Since it is so simple, it can be thought of as a “toy model” (see toy theorem) for theorems on the zeros of analytic functionsMathworldPlanetmath. For a variant of this proof using Rouché’s theorem (which is a consequence of the argument principle) please see the proof of the fundamental theorem of algebra (Rouché’s theorem).

Proof.  Consider the rational functionMathworldPlanetmath

g(z)=zf(z)f(z).

Denote the degree of the polynomialMathworldPlanetmathPlanetmathPlanetmath f by n. Then we can write

g(z)=nzn+lower degree termszn+lower degree terms.

This makes it clear that limzg(z)=n. Hence there exists a real constant R such that |g(z)-n|<1/2 whenever |z|R.

Consider the integral

I=|z|=Rf(z)f(z)𝑑z.

This can be rewritten as

I=|z|=Rg(z)z𝑑z.

Split the integral into two parts, writing I=I1+I2 where

I1=|z|=Rnz𝑑z,I2=|z|=Rg(z)-nz|dz|.

The integral I1 is easy: I1=2πin. As for I2, we shall bound it using our bound for |g(z)-n|.

|I2||z|=R|g(z)-n||z|𝑑z12|dz||z|=π.

Since polynomials are analytic functions in the whole complex planeMathworldPlanetmath, f is an analytic function of z when |z|R, so the argument principle applies and we conclude that I/2πi must equal the number of zeros of f, counted with multiplicity. Among other things, this means that I/2πi must be an integer. By explicit computation, we already know that I1/2πi is also an integer. Hence |I/(2πi)-I1/(2πi)| is an integer. But

|I(2πi)-I1(2πi)|=|I2(2πi)|12.

Now, the only integer smaller that 1/2 in absolute valueMathworldPlanetmathPlanetmathPlanetmath is 0, so we must have I=I1. This implies that f has n zeros (counting with multiplicity) when |z|<R. (By the way we chose R, f(z)0 whenever zR, so f has exactly n zeros in the whole complex plane.)

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更新时间:2025/5/4 6:11:33