How to know if an integral is well defined regardless of path taken. I can calculate 
\begin{equation*}
\int_0^i ze^{z^2} dz=\frac{1}{2e}-\frac12, 
\end{equation*}
but why can I calculate this irrelevant to the path taken? Is this since it is analytic everywhere - if so, how would I go about verifying this? I can't see how to apply the Cauchy Riemann equations here since I don't know how I would break this into the sum of a real and complex component.

Going to the Cauchy-Riemann equations is not a good way of showing that $z\mapsto ze^{z^2}$ is differentiable everywhere, just like reasoning explicitly about $\lim_{h\to 0}\frac{f(x+h)+f(x)}{h}$ wouldn't be a good way to investigate whether the real function $x\mapsto xe^{x^2}$ is differentiable everywhere.
Instead note that the symbolic differentiation rules you learned in ordinary real calculus still work in the complex case: The product and sum of differentiable functions are differentiable, the composition of differentiable functions is differentiable, and so forth -- with the expected derivatives! So
$$ \frac{d}{dz} ze^{z^2} = e^{z^2} + z(\frac d{dz}e^{z^2}) = e^{z^2} + z\cdot 2z \cdot e^{z^2} $$
by the product and rule and then the chain rule. Since this computation works for every $z$, the function $ze^{z^2}$ is differentiable everywhere in the complex plane, and thus analytic, so Cauchy's integral theorem applies to it.
