frame|right|the simple contour c (black), the zeros of f (blue) and the poles of f (red). here we have \oint_{c} {f'(z) \over f(z)}\, dz=2\pi i (4-5). in complex analysis, the argument principle (or cauchy's argument principle) relates the difference between the number of zeros and poles of a meromorphic function to a contour integral of the function's logarithmic derivative. specifically, if f(z) is a meromorphic function inside and on some closed contour c, and f has no zeros or poles on c, then : \oint_{c} {f'(z) \over f(z)}\, dz=2\pi i (n-p) where n and p denote respectively the number of zeros and poles of f(z) inside the contour c, with each zero and pole counted as many times as its multiplicity and order, respectively, indicate. this statement of the theorem assumes that the contour c is simple, that is, without self-intersections, and that it is oriented counter-clockwise. more generally, suppose that f(z) is a meromorphic function on an open set Î© in the complex plane and that c is a closed curve in Î© which avoids all zeros and poles of f and is contractible to a point inside Î©. for each point z â Î©, let n(c,z) be the winding number of c around z. then :\oint_{c} \frac{f'(z)}{f(z)}\, dz = 2\pi i \left(\sum_a n(c,a) - \sum_b n(c,b)\right) where the first summation is over all zeros a of f counted with their multiplicities, and the second summation is over the poles b of f counted with their orders. ==interpretation of the contour integral== the contour integral \oint_{c} \frac{f'(z)}{f(z)}\, dz can be interpreted in two ways: * as the total change in the argument of f(z) as z travels around c, explaining the name of the theorem; this follows from :\frac{d}{dz}\log(f(z))=\frac{f'(z)}{f(z)} and the relation between arguments and logarithms. * as 2&pi;i times the winding number of the path f(c) around the origin, using the substitution w = f(z): :\oint_{c} \frac{f'(z)}{f(z)}\, dz = \oint_{f(c)} \frac{1}{w}\, dw ==proof of the argument principle== let z'n be a zero of f. we can write f(z) = (z&nbsp;&minus;&nbsp;z'n)k'g(z) where k is the multiplicity of the zero, and thus g(z'n) â &nbsp;0. we get : f'(z)=k(z-z_n)^{k-1}g(z)+(z-z_n)^kg'(z)\,\! and : {f'(z)\over f(z)}={k \over z-z_n}+{g'(z)\over g(z)}. since g(z'n) â  0, it follows that g' (z)/g(z) has no singularities at z'n, and thus is analytic at zn, which implies that the residue of f&prim
Theorems_in_complex_analysis