Argument principle

Statement

Let $c \in \mathbb{C}$ and $r > 0$. Let $f$ be a meromorphic function on an open set that contains the closed disc $\{ z : |z - c| \le r \}$. Suppose that $f$ has no zeros and no poles on the circle $|z - c| = r$. Let $Z$ be the number of zeros of $f$ in the open disc $|z - c| < r$, each counted as many times as its order. Let $P$ be the number of poles of $f$ in the open disc, each counted as many times as its order. Then $$\frac{1}{2\pi i} \oint_{|z - c| = r} \frac{f'(z)}{f(z)} \, dz = Z - P,$$ where the circle integral goes once around the circle counterclockwise.

Remarks

The function $f'/f$ is the logarithmic derivative of $f$. At a zero of order $m$, it has a simple pole with residue $m$; at a pole of order $m$, it has a simple pole with residue $-m$. So the theorem follows from the residue theorem. The left side is also the winding number around $0$ of the closed curve $\theta \mapsto f(c + r e^{i\theta})$, $\theta \in [0, 2\pi]$.

Explanations

L1 human

Walk once around the circle and watch the point $f(z)$. Each zero inside the circle makes $f(z)$ turn once around $0$ in the positive direction, and each pole makes it turn once in the negative direction. For example, $f(z) = z^3$ on the unit circle turns three times around $0$, and $f$ has a zero of order $3$ at $0$.