Path independence of complex functions How do i show that $\int e^{-2z}dz$ is independent of the path C joining the points $1-\pi i$ and $2+3\pi i$. Only thing i was able to do was determine the value of this definite integral. Am to to take C to be some arbitary paths like parabolas and straight lines and show that the values are equal? (although that apprach seems stupid.) 

Let $\gamma_1$ and $\gamma_2$ be two paths from $P_1$ to $P_2$ then $\gamma_1\cup \gamma_2^{-}$ is a closed path ($\gamma_2^{-}$ is the path $\gamma_2$ with the orientation reversed). If $f$ is analytic in $\mathbb{C}$ (like your $e^{2z}$), by the Cauchy's Theorem,
$$0=\int_{\gamma_1\cup \gamma_2^{-}} f(z)dz=\int_{\gamma_1} f(z)dz-\int_{\gamma_2} f(z)dz$$
which implies that
$$\int_{\gamma_1} f(z)dz=\int_{\gamma_2} f(z)dz,$$
that is the integral is independent of the path.
To compute the integral, note that $\frac{d}{dz}(e^{2z}/2)=e^{2z}$ and for any path $\gamma$ from $P_1$ to $P_2$,
$$\int_{\gamma} e^{2z}dz=\left [\frac{e^{2z}}{2}\right]_{1−\pi i}^ {2+3\pi i}=\frac{e^{4+6\pi i}-e^{2−2\pi i}}{2}=\frac{e^{4}-e^{2}}{2}.$$
