Argument Principle

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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:

\(\ds \map \LL {\map f z}\) \(=\) \(\ds \frac {\map {f'} z} {\map f z}\) Definition of Logarithmic Derivative of Meromorphic Function
\(\ds \) \(=\) \(\ds \map \LL {\prod_{k \mathop = 1}^N \paren {z - n_k} } - \map \LL {\prod_{j \mathop = 1}^P \paren {z - p_j} } + \map \LL {\map g z}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^N \frac 1 {z - n_k} - \sum_{j \mathop = 1}^P \frac 1 {z - p_j} + \frac {\map {g'} z} {\map g z}\)

Then:

\(\ds \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z\) \(=\) \(\ds \oint_\gamma \paren {\sum_{k \mathop = 1}^N \frac 1 {z - n_k} - \sum_{j \mathop = 1}^P \frac 1 {z - p_j} + \frac {\map {g'} z} {\map g z} } \rd z\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^N \oint_\gamma \frac 1 {z - n_k} \rd z - \sum_{j \mathop = 1}^P \oint_\gamma \frac 1 {z - p_j} \rd z + \oint_\gamma \frac {\map {g'} z} {\map g z} \rd z\) Linear Combination of Contour Integrals
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^N 2 \pi i - \sum_{j \mathop = 1}^P 2 \pi i + \oint_\gamma \frac {\map {g'} z} {\map g z} \rd z\)
\(\ds \) \(=\) \(\ds 2\pi i \paren {N - P} + \oint_\gamma \frac {\map {g'} z} {\map g z} \rd z\)

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:

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

$\blacksquare$