I'm trying to show that if $f(z)$ is analytic an $|f(z) - 1| < 1$, then $\int_\gamma \frac{f'(z)}{f(z)}dz = 0$ over all closed curves $\gamma$.
Presumably, I need to show that this is an exact differential. I can rewrite $\frac{f'(z)}{f(z)}dz$ as
$$\frac{f'(z)}{f(z)}dz = \left(\frac{1}{u + iv} \frac{\partial u}{\partial x} + \frac{v + iu}{u^2+v^2}\frac{\partial v}{\partial x}\right)(dx + idy);$$
however, now I need to find some $F(z)$ such that the partial derivatives of $F$ form the expression above (multiplied by $i$ for the partial on $y$). My idea was to show that both the real and imaginary parts of the differential are exact to simplify it a bit, but my algebraic manipulations don't seem to go anywhere. Is there something I'm missing; perhaps the $|f(z) - 1|<1$ condition? (For now, I'm assuming that that's just there so the denominator is not zero).