Analytical region of complex function $\frac{1}{z^3+a}$ ## Original Problem: For complex function $f(z)$: $$\frac{1}{z^3+a}$$ with $a \in \mathbb R, a > 0$ Find its region of analyticity, as well as its derivative. ## My question: I have trouble finding its region of analyticity, i.e. where the function is differentiable, or satisfying C-R equation. It is apparently unwise to let $z = x + iy$ and expand the function, **but I don't know How to use $z$ directly**. ## My assumption without any proof: Except for the only point $z = -\sqrt[3]{a}$, $f(z)$ is analytical in the whole complex plane. The derivative can be calculated as if $z$ is real number in real function, which leads to $$f'(z) = \frac{-3z^2}{(z^3+a)^2}$$

The function is a composition of analytic functions $z \mapsto z^3 + a$ and $w \mapsto \frac{1}{w}$, so it is analytic wherever the composition is defined. The first function is defined everywhere, and the second is defined everywhere except at $w = 0$, so the composition is analytic everywhere except where $z^3 + a = 0$.

One solution of this equation is (almost) the one you found, the real number $-\sqrt[3]{a}$. Notice that if $\omega$ is a third root of unity, say $\omega = e^{2\pi i / 3}$, then $(\omega z)^3 = \omega^3 z^3 = z^3$, and so $\omega(-\sqrt[3]{a})$ and $\omega^2(-\sqrt[3]{a})$ are also solutions of $z^3 + a$. The equation is cubic, so the three solutions we've found are all of the solutions, and therefore all of the points in $\mathbb{C}$ where the function is not analytic. (This technique works just as well solving equations of the form $z^k + b$ for positive integers $k$, by the way.)

The derivative computation is correct.