# Moving the branch cut of the complex logarithm

The complex logarithm is defined as $\log z:=\operatorname{Log} |z|+i\arg z$ , with the branch cut on the non-negative real axis.

Determine a branch of $f(z)=\log(z^3-2)$ that is analytic at $z=0$ and find $f(0)$ and $f'(0)$.

So, I would need to choose a branch of the logarithm such that it is analytic at $g(0)=-2$ (where $g(z)=z^3-2)$. I am having difficulty knowing which branch can be chosen. The book I am using claims that $\mathcal{L_{-\pi/4}}$ works, and shows that the branch cut has moved from the negative real axis to somewhere in quadrant $4$. Why is this? I would have thought that the branch cut would have been moved to somewhere in quadrant $2$ if I moved it by $-\pi/4$.

• Taking a single-valued branch is always annoying to me. +1 – Yai0Phah Apr 19 '14 at 14:25
• It works if we rewrite $f(z)=\log(z-\sqrt[3]2)+\log(z-\sqrt[3]2\omega)+\log(z-\sqrt[3]2\omega^2)$, but I want to obtain good insight on how Riemann surfaces work. For example, what's the exact meaning of $\log(fg)=\log f+\log g$ w.r.t. Riemann surfaces when $f,g$ are multi-valued functions. – Yai0Phah Apr 19 '14 at 14:28