Argument Principle

Theorem
Let $\gamma$ be a closed contour.

Let $D$ be the region enclosed by $\gamma$.

Let $f$ be a function meromorphic on $D$.

Let $f$ be holomorphic with no zeroes on $\gamma$.

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

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

Then:
 * $\ds N - P = \frac 1 {2\pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$

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 non-zero holomorphic function $g$ such that:


 * $\ds \map f x = \frac {\prod_{k \mathop = 1}^N \paren {z - n_k} } {\prod_{j \mathop = 1}^P \paren {z - p_j} } \map g z$

Taking the logarithmic derivative:

Then:

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

We therefore have:


 * $\ds \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z = 2\pi i \paren {N - P}$

Hence:


 * $\ds N - P = \frac 1 {2\pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$