Question: What is the anti-derivative of $z^{i}$?
Motivation: While doing some complex analysis problems, I got to one which required me to find the anti-derivative of $z^{i}$. In the solutions, it notes (without explanation) that the anti-derivative is $z^{i+1}$ (of course!). I think this is in error (since, in particular, taking the derivative of the expanded form $e^{(i+1)\log(z)}$ gives $\frac{(i+1)}{z}e^{(i+1)\log(z)} = (i+1)e^{i\log(z)}$, but I wanted to make sure I wasn't making any silly mistakes. I also plugged this into Wolfram Alpha but whenever complex things come into play WA has a tendency (at least for me) to give some strange answers sometimes (especially when branch cuts and things come up).
Further (Optional!) Question: Is there a power formula for the antiderivative of $z^{a+bi}$? The proof of the real exponent case usually requires the binomial theorem, and I don't know a similar theorem for complex numbers.