Argument Principle

Theorem
Let $f$ be a function meromorphic in the interior of some simply connected region $D$.

Let $f$ be holomorphic with no zeroes on the boundary of $D$.

Let $N$ denote the number of zeroes of $f$ in the interior of $D$, counted up to multiplicity.

Let $P$ denote the number of poles of $f$ in the interior of $D$, counted up to order.

Then:
 * $\displaystyle N - P = \frac 1 {2\pi i} \oint_D \frac { f'\left({z}\right) } { f\left({z}\right) } \, \mathrm d z$

Proof
Let $n_1, n_2, n_3 \ldots n_N$ denote the zeroes of $f$, and $p_1, p_2, p_3 \ldots p_P$ denote its poles. Then, there exists a function $g$ nonzero and holomorphic on the boundary and on the interior of $D$ such that:


 * $\displaystyle f\left({z}\right) = \frac { \prod_{k \mathop = 1}^N \left( { z - n_k } \right) } { \prod_{j \mathop = 1}^P \left( { z - p_j } \right) } g\left({z}\right)$

Taking the logarithmic derivative:

Then:

By our construction of $g$, $\frac {g'} g$ is holomorphic on and inside $D$ so, by the Cauchy Integral Theorem, the integral is equal to zero. We therefore have:


 * $\displaystyle \oint_D \frac { f'\left({z}\right) } { f\left({z}\right) } \, \mathrm d z = 2\pi i \left( {N - P} \right)$

Hence:


 * $\displaystyle N - P = \frac 1 {2\pi i} \oint_D \frac { f'\left({z}\right) } { f\left({z}\right) } \, \mathrm d z$