# Taking the anti-derivative of $z^{i}$

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.

• Do you mean $z^{i+1}/(i+1)$ rather than $z^{i+1}$? Commented Aug 4, 2011 at 21:13
• That is what I suspected, and was hoping the book was in error for leaving out!
– user2959
Commented Aug 5, 2011 at 18:21

Yes, the notes must have left out the constant. The anti-derivative is of course $z^{i+1} / (i+1)$, just as in the real case. A proof can be done just by noting the definition of complex powers in terms of exponentials and logarithms (just as you did). Do note all the needed disclaimers about multiplicity, staying on the same Riemann surface, not crossing branch points and cuts, and so forth, and the special case of $-1$.