Answer: Logarithmic Derivative and Integrality of Winding

Key Idea / Intuition

The idea is that $\log f(z)$, while not globally well-defined (since $\log$ is multi-valued), has a perfectly well-defined derivative: $(\log f)' = f'/f$. Integrating this derivative around a closed loop measures how much the argument of $f$ winds around, which must be an integer multiple of $2\pi$ — because $f$ returns to its starting value. The integral $\frac{1}{2\pi i}\oint f'/f\, dz$ is literally counting how many times $f(\gamma)$ winds around the origin.

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Formal Proof / Solution

Step 1: The warm-up — $f(z) = z^n$

As $z = e^{i\theta}$ with $\theta$ going from $0$ to $2\pi$, we have $f(z) = e^{in\theta}$, so $\arg f = n\theta$. The total change in argument is: $\Delta \arg f = n \cdot 2\pi.$

Step 2: Setting up the integral

Suppose $f$ is analytic and nonvanishing on a closed curve $\gamma : [0,1] \to \mathbb{C}$. We want to compute: $I = \frac{1}{2\pi i} \oint_\gamma \frac{f'(z)}{f(z)}\, dz.$

Step 3: Local logarithm trick

Along the curve $\gamma$, since $f(\gamma(t)) \neq 0$ for all $t \in [0,1]$, we can define a continuous branch of the logarithm along the curve. That is, there exists a continuous function $L : [0,1] \to \mathbb{C}$ such that: $e^{L(t)} = f(\gamma(t)), \quad L(t) = \ln|f(\gamma(t))| + i\,\arg f(\gamma(t)).$

(This is the monodromy/path-lifting theorem for the exponential map, or simply follows by continuity: locally, $\log f$ is analytic since $f \neq 0$.)

Step 4: The integral reduces to boundary values

By the chain rule and the substitution $z = \gamma(t)$: $\oint_\gamma \frac{f'(z)}{f(z)}\, dz = \int_0^1 \frac{f'(\gamma(t))}{f(\gamma(t))}\, \gamma'(t)\, dt = \int_0^1 \frac{d}{dt} L(t)\, dt = L(1) - L(0).$

Step 5: Integrality

Since $\gamma$ is closed, $f(\gamma(1)) = f(\gamma(0))$, which means $e^{L(1)} = e^{L(0)}$. Therefore: $L(1) - L(0) = 2\pi i \cdot k$ for some integer $k \in \mathbb{Z}$.

Thus: $\frac{1}{2\pi i} \oint_\gamma \frac{f'(z)}{f(z)}\, dz = k \in \mathbb{Z}.$

Step 6: Geometric meaning

The integer $k$ is the winding number of the image curve $f \circ \gamma$ around $0$. In the warm-up example $f(z) = z^n$ on the unit circle: $\frac{1}{2\pi i} \oint_{|z|=1} \frac{nz^{n-1}}{z^n}\, dz = \frac{1}{2\pi i} \cdot \frac{n}{z}\, dz = n.$

This confirms the winding is exactly $n$.

Bonus: Connection to the Argument Principle

When $f$ may have zeros and poles inside $\gamma$, the same reasoning gives the Argument Principle: $\frac{1}{2\pi i} \oint_\gamma \frac{f'(z)}{f(z)}\, dz = Z - P,$ where $Z$ = number of zeros and $P$ = number of poles of $f$ inside $\gamma$ (counted with multiplicity). The integrality is automatic from the winding number perspective.