I would appreciate it if someone checked my work to ensure that it's consistent.
Compute the integral $\int_{C}\frac 1 z {dz}$ by obtaining an appropriate branch of the logarithm.
There's an image provided showing an unspecified point in quadrant III creating a line towards another unspecified point in quadrant II.
Let the branch of $\text{Log}(z)$ be defined as $0 \lt \arg(z) \lt 2\pi$, which implies that the branch is defined by the positive real axis.
$\int_{C}\frac 1 z {dz}$ where $z\neq 0$ and $ z\not\in\Bbb R^+$
Then for some starting point $z_i$ in quadrant III and some ending point $z_o$ in quadrant II, $z$ can be parametrized:
$x = x_i + x_ot$
$y = y_i + y_ot$
$z = x_i + x_ot + i(y_i + y_ot) = z_i + z_ot$, where $0 \le t \le 1$
$dz = z_odt$
$\int_0^1 \frac {1} {z_i + z_o} z_odt = [\text{Log}(z_i + z_ot)]_0^1 = \text{Log}(z_i + z_o) - \text{Log}(z_i)$
Would someone mind explaining the topology of this as well? I've always assumed that line integrals were path independent. If the branch cut was defined by $-\pi \lt \arg(z) \lt \pi$, could an arc parametrization be created to integrate around the branch cut from $z_i$ to $z_o$? This isn't what the assignment asked, but I'm curious. Thanks!