Proof of The Argument Principle
Let zN be a zero of f. We can write f(z) = (z − zN)kg(z) where k is the multiplicity of the zero, and thus g(zN) ≠ 0. We get
and
Since g(zN) ≠ 0, it follows that g' (z)/g(z) has no singularities at zN, and thus is analytic at zN, which implies that the residue of f′(z)/f(z) at zN is k.
Let zP be a pole of f. We can write f(z) = (z − zP)−mh(z) where m is the order of the pole, and thus h(zP) ≠ 0. Then,
and
similarly as above. It follows that h′(z)/h(z) has no singularities at zP since h(zP) ≠ 0 and thus it is analytic at zP. We find that the residue of f′(z)/f(z) at zP is −m.
Putting these together, each zero zN of multiplicity k of f creates a simple pole for f′(z)/f(z) with the residue being k, and each pole zP of order m of f creates a simple pole for f′(z)/f(z) with the residue being −m. (Here, by a simple pole we mean a pole of order one.) In addition, it can be shown that f′(z)/f(z) has no other poles, and so no other residues.
By the residue theorem we have that the integral about C is the product of 2πi and the sum of the residues. Together, the sum of the k 's for each zero zN is the number of zeros counting multiplicities of the zeros, and likewise for the poles, and so we have our result.
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