Argument Principle - Proof of The Argument Principle

Proof of The Argument Principle

Let z_N be a zero of f. We can write f(z) = (z − z_N)kg(z) where k is the multiplicity of the zero, and thus g(z_N) ≠ 0. We get

and

Since g(z_N) ≠ 0, it follows that g' (z)/g(z) has no singularities at z_N, and thus is analytic at z_N, which implies that the residue of f′(z)/f(z) at z_N is k.

Let z_P be a pole of f. We can write f(z) = (z − z_P)−mh(z) where m is the order of the pole, and thus h(z_P) ≠ 0. Then,

and

similarly as above. It follows that h′(z)/h(z) has no singularities at z_P since h(z_P) ≠ 0 and thus it is analytic at z_P. We find that the residue of f′(z)/f(z) at z_P is −m.

Putting these together, each zero z_N of multiplicity k of f creates a simple pole for f′(z)/f(z) with the residue being k, and each pole z_P 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 z_N is the number of zeros counting multiplicities of the zeros, and likewise for the poles, and so we have our result.