[proofplan] Since $f$ has no zeros on $\gamma$ and $\gamma$ is compact, $|f|$ attains a positive minimum $\delta > 0$ on $\gamma$. Uniform convergence on $\gamma$ guarantees $|f_n - f| < \delta$ on $\gamma$ for all sufficiently large $n$, which forces $f_n$ to be nonzero on $\gamma$. The difference of the two winding-number integrals equals the winding number of $f_n/f \circ \gamma$ around the origin, and the estimate $|f_n/f - 1| < 1$ on $\gamma$ confines this image to the half-plane $\operatorname{Re}(w) > 0$, forcing the winding number to be zero. [/proofplan] [step:Establish a positive lower bound for $|f|$ on $\gamma$] Since $\gamma$ is a compact subset of $\Omega$ and $f$ is continuous on $\Omega$ with no zeros on $\gamma$, the function $|f|$ attains its minimum on $\gamma$. Define \begin{align*} \delta := \min_{z \in \gamma} |f(z)| > 0. \end{align*} The strict positivity follows because $f(z) \neq 0$ for all $z \in \gamma$ by hypothesis. [guided] Why does this step matter? The entire argument rests on showing that $f_n$ is eventually nonzero on $\gamma$, which requires a quantitative gap between $|f|$ and $0$ on $\gamma$. Since $\gamma$ is compact (it is the image of a closed bounded interval under a continuous parametrisation) and $|f|: \gamma \to \mathbb{R}$ is continuous (as the composition of the holomorphic function $f$ with the continuous map $|\cdot|$), the extreme value theorem guarantees that $|f|$ attains its infimum on $\gamma$. Since $f$ has no zeros on $\gamma$ by hypothesis, this minimum value is strictly positive. We set \begin{align*} \delta := \min_{z \in \gamma} |f(z)| > 0. \end{align*} This $\delta$ will serve as the threshold: once $\|f_n - f\|_\gamma < \delta$, no cancellation can occur on $\gamma$. [/guided] [/step] [step:Show $f_n$ has no zeros on $\gamma$ for all sufficiently large $n$] The trace of $\gamma$ is a compact subset of $\Omega$. Since $f_n \to f$ uniformly on compact subsets of $\Omega$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$, \begin{align*} \max_{z \in \gamma} |f_n(z) - f(z)| < \delta. \end{align*} For any $z \in \gamma$ and $n \geq N$, the reverse triangle inequality gives \begin{align*} |f_n(z)| \geq |f(z)| - |f_n(z) - f(z)| > \delta - \delta = 0. \end{align*} Hence $f_n(z) \neq 0$ for all $z \in \gamma$ and all $n \geq N$, so the integrals $\frac{1}{2\pi i}\oint_\gamma \frac{f_n'(z)}{f_n(z)}\,dz$ are well-defined for $n \geq N$. [guided] We need both integrals in the theorem statement to be well-defined, which requires $f_n$ to be nonzero on $\gamma$. The key tool is uniform convergence on the compact set $\gamma$. Since $f_n \to f$ uniformly on compact subsets of $\Omega$ and the trace of $\gamma$ is compact (as the continuous image of a closed interval), there exists $N \in \mathbb{N}$ such that \begin{align*} \sup_{z \in \gamma} |f_n(z) - f(z)| < \delta \quad \text{for all } n \geq N. \end{align*} Now fix any $z \in \gamma$ and any $n \geq N$. By the reverse triangle inequality: \begin{align*} |f_n(z)| = |f(z) + (f_n(z) - f(z))| \geq |f(z)| - |f_n(z) - f(z)| \geq \delta - \delta = 0. \end{align*} But the inequality is strict: $|f_n(z) - f(z)| < \delta \leq |f(z)|$, so $|f_n(z)| > |f(z)| - |f_n(z) - f(z)| > 0$. Therefore $f_n$ has no zeros on $\gamma$ for $n \geq N$, and the logarithmic derivative $f_n'/f_n$ is holomorphic in a neighbourhood of $\gamma$. [/guided] [/step] [step:Express the difference of winding-number integrals as a single contour integral] For $n \geq N$, define the quotient $h_n: \Omega \to \mathbb{C}$ by $h_n(z) = f_n(z)/f(z)$. Since $f$ has only isolated zeros in $\Omega$ (being holomorphic and not identically zero on any connected component containing $\gamma$) and has no zeros on $\gamma$, the function $h_n$ is holomorphic in a neighbourhood of $\gamma$. The logarithmic derivative satisfies \begin{align*} \frac{h_n'(z)}{h_n(z)} = \frac{f_n'(z)}{f_n(z)} - \frac{f'(z)}{f(z)}, \end{align*} which is obtained by differentiating $\log h_n = \log f_n - \log f$ (valid locally since neither $f_n$ nor $f$ vanishes on $\gamma$). Therefore \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{f_n'(z)}{f_n(z)}\,dz - \frac{1}{2\pi i}\oint_\gamma \frac{f'(z)}{f(z)}\,dz = \frac{1}{2\pi i}\oint_\gamma \frac{h_n'(z)}{h_n(z)}\,dz = n(h_n \circ \gamma, 0), \end{align*} where $n(h_n \circ \gamma, 0)$ denotes the winding number of the closed curve $h_n \circ \gamma$ around the origin. [guided] The idea is to reduce the equality of two integrals to showing that a single winding number vanishes. Consider the quotient $h_n = f_n / f$. Since $f$ is holomorphic and nonzero on $\gamma$, and $f_n$ is holomorphic on $\Omega$, the function $h_n$ is holomorphic in a neighbourhood of $\gamma$. By the quotient rule for logarithmic derivatives (which follows from differentiating $f_n = h_n \cdot f$ and dividing by $f_n = h_n \cdot f$): \begin{align*} \frac{f_n'(z)}{f_n(z)} = \frac{(h_n \cdot f)'(z)}{(h_n \cdot f)(z)} = \frac{h_n'(z) f(z) + h_n(z) f'(z)}{h_n(z) f(z)} = \frac{h_n'(z)}{h_n(z)} + \frac{f'(z)}{f(z)}. \end{align*} Rearranging and integrating over $\gamma$: \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{f_n'(z)}{f_n(z)}\,dz - \frac{1}{2\pi i}\oint_\gamma \frac{f'(z)}{f(z)}\,dz = \frac{1}{2\pi i}\oint_\gamma \frac{h_n'(z)}{h_n(z)}\,dz. \end{align*} The right-hand side is the winding number of the image curve $h_n \circ \gamma$ around the origin (by the definition of winding number via the logarithmic derivative). So to prove equality of the two original integrals, it suffices to show $n(h_n \circ \gamma, 0) = 0$. [/guided] [/step] [step:Show the image of $h_n \circ \gamma$ lies in a half-plane not containing the origin] For $z \in \gamma$ and $n \geq N$: \begin{align*} |h_n(z) - 1| = \left|\frac{f_n(z)}{f(z)} - 1\right| = \frac{|f_n(z) - f(z)|}{|f(z)|} \leq \frac{|f_n(z) - f(z)|}{\delta} < \frac{\delta}{\delta} = 1. \end{align*} Therefore the image $h_n(\gamma) \subset B(1, 1) \subset \{w \in \mathbb{C} : \operatorname{Re}(w) > 0\}$. Since the open disc $B(1,1)$ is a convex set that does not contain the origin (as $|0 - 1| = 1$ and the disc is open), the winding number of $h_n \circ \gamma$ around the origin is zero: \begin{align*} n(h_n \circ \gamma, 0) = 0. \end{align*} [guided] We need to show the winding number $n(h_n \circ \gamma, 0) = 0$. The strategy is to show that the image curve $h_n \circ \gamma$ stays in a region that does not wind around the origin. For any $z \in \gamma$ and $n \geq N$, we estimate: \begin{align*} |h_n(z) - 1| = \left|\frac{f_n(z) - f(z)}{f(z)}\right| = \frac{|f_n(z) - f(z)|}{|f(z)|}. \end{align*} Since $|f(z)| \geq \delta$ (from Step 1) and $|f_n(z) - f(z)| < \delta$ (from Step 2), we obtain \begin{align*} |h_n(z) - 1| < \frac{\delta}{\delta} = 1. \end{align*} This means the image $h_n(\gamma)$ is contained in the open disc $B(1, 1) = \{w \in \mathbb{C} : |w - 1| < 1\}$. Why does this imply the winding number is zero? The disc $B(1,1)$ does not contain the origin (since $|0 - 1| = 1$ and the disc is open). Moreover, $B(1,1)$ is convex, hence simply connected. A closed curve lying entirely in a simply connected region that does not contain a point $w_0$ necessarily has winding number zero around $w_0$. (Alternatively: $B(1,1) \subset \{w : \operatorname{Re}(w) > 0\}$, and any curve in the right half-plane has a continuous argument, hence zero winding number around the origin.) Therefore \begin{align*} n(h_n \circ \gamma, 0) = 0. \end{align*} [/guided] [/step] [step:Conclude equality of the zero-counting integrals] Combining the results of Steps 3 and 4, for all $n \geq N$: \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{f_n'(z)}{f_n(z)}\,dz - \frac{1}{2\pi i}\oint_\gamma \frac{f'(z)}{f(z)}\,dz = n(h_n \circ \gamma, 0) = 0. \end{align*} By the [Argument Principle](/theorems/???), each integral counts the number of zeros (with multiplicity) of the respective function inside $\gamma$. Since both integrals are integers and they are equal for $n \geq N$, the number of zeros of $f_n$ inside $\gamma$ equals the number of zeros of $f$ inside $\gamma$ (counted with multiplicity) for all sufficiently large $n$. [/step]