I have been studying complex integration and I have been calculating some simple contour integrals. By integrating the function $f(z) = \frac{1}{z}$ around the unit circle centered on the origin I get $\int_{C} = 2\pi i$ as expected by the residue theorem.
I then repeated using a different countour, a circle of radius 1 centered on $z=3$, $g(t) = 3 + e^{it}$ where $0 < t < 2\pi$ where I get an integral of zero, again as predicted by the residue theorem. (The singularity at $z=0$ is not inside this contour integral).
I then tried to integrate around the circle with center $z=0$ and radius 4. I was expecting/hoping to get the value $2\pi i$ since the function $f(z) = \frac{1}{z}$ has a single singularity included inside it's contour. Although this is probably the most simple application of the residue theorem that anyone applies when first studying the topic, however, I get an integral of zero. Surely the residue theorem would dictate a value of $2\pi i$ around this countour as it includes the one and only singularity at $z = 0$? I expect that I am misunderstanding some subtlety of the logarithm function. I have included my working below:
$$f(z) = \frac{1}{z}$$ $$g(t) = 3 + 4e^{it} \quad 0 < t < 2\pi$$ $$\int_{g}f(g(t))g'(t)dt = \int_{0}^{2\pi}\dfrac{4ie^{it}}{3+4e^{it}}dt = \ln(3+4e^{2\pi i}) - \ln(3 + 4e^{0}) = \ln(7) - \ln(7) = 0$$
Recall that the complex log function is branched, with each branch corresponding to a different branch of the argument function: $\log(re^{i\theta}) = \ln(r) + i\theta,$ so the value depends on our choice of $\theta.$
If we’re integrating for $\theta$ going from $0$ to $2\pi,$ then it’s most natural for us to pick $0 \leq \theta < 2\pi,$ so let’s go with that here. In that case, since this branch doesn’t consider $\theta = 2\pi$ as part of its domain, technically we can consider our integral as an improper integral since one of the limits isn’t in the domain. So, our integral should be
$$\lim_{b \to 2\pi^-} \log(3 + 4e^{ib}) - \ln(7)$$
so, considering that
$$z_b = 3 + 4e^{ib} = (3 + 4\cos b) + 4i\sin b$$
so
$$|z_b| = \sqrt{25 + 24\cos b} \\ \tan(\arg(z_b)) = \frac{4\sin b}{3 + 4\cos b}$$
and since $z_b$ is in the fourth quadrant for the $b$ we’ll be considering near $2\pi,$ in particular we would have
$$\log(z_b) = \ln(\sqrt{25 + 24e^{ib}}) + i\left(2\pi + \arctan\left( \frac{4\sin b}{3 + 4\cos b}\right)\right)$$
and taking the limit it should be clear that $$\lim_{b \to 2\pi^-} \log(z_b) - \ln(7) = 2\pi i$$
matching the answer from the residue theorem.
Hope this helps!