why is this definite complex integral path independant? in a book it is said that this integral is path independent
$$\int_{i\pi}^{3i\pi} e^{3z}dz$$
if it's the case then computing the line integral $\int_{\gamma} e^{3z}dz$ where $\gamma$ is the straight line between $i\pi$ and $3i\pi$ makes life easy but why choosing any path/curve doesn't change the value of this integral ?
 A: A simple approach to the problem is to note that $e^{3z}$ has an antiderivative. Therefore, if $\gamma : [0,1]\rightarrow\mathbb{C}$ is a continuous curve of  bounded variation from $i\pi$ to $3i\pi$, then
\begin{align}
     \int_{i\pi}^{3i\pi}e^{3z}dz &=\int_{0}^{1}e^{3\gamma(t)}d\gamma(t) \\
       & =\int_{0}^{1}d_{t}\left(\frac{1}{3}e^{3\gamma(t)}\right) \\
       & =\left.\frac{1}{3}e^{\gamma(t)}\right|_{t=0}^{1} \\
       & = \frac{1}{3}\left(e^{3\pi i}-e^{\pi i}\right)
\end{align}
The answer does not depend on the particular choice of continuous curve $\gamma$ as described.
A: Hint: The function $e^{3z}$ is holomorphic for all $\mathbb{C}$, you can check this by using the Cauchy-Riemann equations. By the Cauchy Integral theorem it follows that $\int_{\gamma}e^{3z}dz$ is path indendent, as long as the starting point and the end point of your path stay the same.
A: Because the function $f(z) = e^{3z}$ is analytic everywhere, and hence, but the Cauchy Integral formula, its integral over any closed loop is zero. 
