Editing Argument principle (section)

==Proof of the argument principle==
Let ''z''<sub>''Z''</sub> be a zero of ''f''. We can write ''f''(''z'') = (''z''&nbsp;&minus;&nbsp;''z''<sub>''Z''</sub>)<sup>''k''</sup>''g''(''z'') where ''k'' is the multiplicity of the zero, and thus ''g''(''z''<sub>''Z''</sub>) ≠&nbsp;0. We get

: <math>f'(z)=k(z-z_Z)^{k-1}g(z)+(z-z_Z)^kg'(z)\,\!</math>

and

: <math>{f'(z)\over f(z)}={k \over z-z_Z}+{g'(z)\over g(z)}.</math>

Since  ''g''(''z''<sub>''Z''</sub>) ≠ 0, it follows that ''g' ''(''z'')/''g''(''z'') has no singularities at ''z''<sub>''Z''</sub>, and thus is analytic at ''z''<sub>Z</sub>, which implies that the [[Residue (complex analysis)|residue]] of ''f''&prime;(''z'')/''f''(''z'') at ''z''<sub>''Z''</sub> is&nbsp;''k''.

Let ''z''<sub>P</sub> be a pole of ''f''. We can write ''f''(''z'') = (''z''&nbsp;&minus;&nbsp;''z''<sub>P</sub>)<sup>&minus;''m''</sup>''h''(''z'') where ''m'' is the order of the pole, and 
''h''(''z''<sub>P</sub>) ≠ 0. Then,

: <math>f'(z)=-m(z-z_P)^{-m-1}h(z)+(z-z_P)^{-m}h'(z)\,\!.</math>

and

: <math>{f'(z)\over f(z)}={-m \over z-z_P}+{h'(z)\over h(z)}</math>

similarly as above. It follows that ''h''&prime;(''z'')/''h''(''z'') has no singularities at ''z''<sub>P</sub> since ''h''(''z''<sub>P</sub>) ≠ 0 and thus it is analytic at ''z''<sub>P</sub>. We find that the residue of
''f''&prime;(''z'')/''f''(''z'') at ''z''<sub>P</sub> is &minus;''m''.

Putting these together, each zero ''z''<sub>''Z''</sub> of multiplicity ''k'' of ''f'' creates a simple pole for
''f''&prime;(''z'')/''f''(''z'') with the residue being ''k'', and each  pole ''z''<sub>P</sub> of order ''m'' of
''f'' creates a simple pole for ''f''&prime;(''z'')/''f''(''z'') with the residue being &minus;''m''. (Here, by a simple pole we
mean a pole of order one.) In addition, it can be shown that ''f''&prime;(''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 2''πi'' and the sum of the residues.  Together, the sum of the ''k''{{'}}s for each zero ''z''<sub>''Z''</sub> is the number of zeros counting multiplicities of the zeros, and likewise for the poles, and so we have our result.