Integral yields log of holomorphic function
I will fill in Alex Youcis's suggestion. So, By the argument principle, the integral in question is equal to $2\pi i($number of zeros - number of poles). Since $f$ is holomorphic in the region, there are no poles. So there are only zeros. By Rouche, $f(z)$ has as many zeros as the function 1, i.e. zero.