The integral of $f'/f$ along a loop is exactly the winding number of its image around $0$.Suggest a slogan
Let \(\Omega\subseteq\CC\) be open, let \(f\) be meromorphic on \(\Omega\), and let \(\gamma\colon[a,b]\to\Omega\) be a piecewise smooth closed curve on which \(f\) has no zeros or poles. Then \[ \frac{1}{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\dz=\Index_{w=0}(f\circ\gamma), \] the winding number of the closed curve \(f\circ\gamma\) about \(0\).
The curve \(f\circ\gamma\) is closed, piecewise smooth, and avoids \(0\). Substituting \(w=f(\gamma(t))\), so that \(\dw=f'(\gamma(t))\gamma'(t)\dt\), gives \[ \Index_{w=0}(f\circ\gamma)=\frac{1}{2\pi i}\int_{f\circ\gamma}\frac{\dw}{w}=\frac{1}{2\pi i}\int_a^b\frac{f'(\gamma(t))}{f(\gamma(t))}\gamma'(t)\dt=\frac{1}{2\pi i}\int_\gamma\frac{f'(z)}{f(z)}\dz. \]