Use the Fundamental Theorem of Calculus (FTC) to calculate $\int_a^b \frac{1}{z} dz$ for any pair of non-zero complex numbers $a,b$ such that the line segment from $a$ to $b$ does not contain $0$. Then, use your answer to find an example of a triangle $\Delta$ with vertices $a,b,c$ such that $\int_a^c \frac{1}{z} dz - \int_a^b \frac{1}{z} dz \neq \int_b^c \frac{1}{z} dz$.
Here's what I have so far: By FTC, we know that $\int_{\gamma} f(z)dz = g(\gamma(b))-g(\gamma(a))$. We also know that $f(z)=\frac{1}{z}$ has an antiderivative given by $\log(z)$ away from the cut line.
I am thus tempted to conclude that $\int_a^b \frac{1}{z} dz = \log(b) - \log(a)$, but this wouldn't seem to be conducive to finding an example for the second part of the problem.
Any help would be greatly appreciated. Thanks.
The problem is that the "cut line" for the $\log$ function has to change depending on where $a$ and $b$ are. The cut line cannot intersect the line segment from $a$ to $b$, since the antiderivative has to be defined on the entire curve of integration.
Given particular points $a$ and $b$, you can find a cut line that works, but there is not a single cut line that works for all $a$ and $b$. Equivalently, there is no single function "$\log$" that works for all $a$ and $b$.
This should help you find a counterexample for the second part of the problem.