I am trying to find the value of $\int_{C} \frac{1}{z} dz$, where C is the path from $-1-i$ to $-1+i$. I have computed this integral in two different ways. The first method was computing the integral directly, which gave me a result of $\frac{-\pi}{2}i$. The second method, which was knowing that $F(z) = \ln (z)$ is a primitive function (a possible result of integration) and evaluating $F(-1+i)-F(-1-i)$, which gives me a result of $\frac{3\pi}{2}i$. Therefore, I'm a bit puzzled of which one is the correct answer. I suspect it's the second method, since it's based on the fundamental theorem of calculus, otherwise I'm not sure and I'd like to back that up with better arguments.
Any help would be appreciated, Thank you!
If your path is the straight line from $-1 - i$ to $-1 + i$, then the correct answer is $-\frac{\pi}{2}i$.
When you use the second method, you must choose a branch of $\log z$ defined in a domain that contains your path. Here, you can use the following branch: $$\log z = \log|z| + i \operatorname{arg} z,$$ where $0 \leq \operatorname{arg} z < 2\pi$. This is defined in the slit domain $\mathbb{C} \backslash [0, +\infty)$. With this branch, we have $\log(-1 - i) = \log \sqrt{2} + i \frac{5\pi}{4}$, $\log (-1 + i) = \log \sqrt{2} + i \frac{3\pi}{4}$, and $$\int_C \frac{1}{z}dz = \log(-1 + i) - \log(-1 - i) = -i\frac{\pi}{2}.$$