Analytic continuation of the function $f=\sqrt[3]{z^2-1}$

Question

We consider analytic continuation of the function $f=\sqrt[3]{z^2-1}$ along the semicircle $|z| = 3$ in the lower half plane. We know that $f(3) = 2$. How can one find $f(-3)$?

Answer 1

The functions $u(z) = \sqrt[3]{1+z}$ and $v(z) = \sqrt[3]{1-z}$ are analytic on the lower half plane, so their product $u(z)v(z)=\sqrt[3]{1-z}\sqrt[3]{1+z}$ is also analytic in the lower half plane, and it agrees with $f(z)$ near $3$, so you may calculate the value at $-3$ as $$\lim_{y \to 0^-}u(-3+y\, i) v(-3+y\,i) = -1-i\sqrt{3}$$