# I don't get branch cuts in the complex plane at ALL

1. Sep 3, 2009

### AxiomOfChoice

Suppose you're trying to provide a branch cut in $\mathbb{C}$ that will define a single-value branch of $f(z) = \log(z - z_0)$. I don't know where to begin. Can someone help explain this concept to me?

2. Sep 4, 2009

### HallsofIvy

Staff Emeritus
You can first shift to the origin by letting u= z- z0 so your question refers to f(u)= ln(u). Now any straight line, from 0 to infinity can be taken as a branch cut! The "standard" choice is to take the negative x-axis as branch cut so that $ln(z)= ln(re^{i\theta})= ln(r)+ i\theta$ and $-\pi< \theta< \pi$. The point is that that choice of branch cut "separates" $-\pi< \theta< \pi$ from $\pi< \theta< 3\pi$ from $3\pi< \theta< 5\pi$, etc.

In terms of your original z, a branch cut is a straight line from z0 to infinity and the "standard" choice is the line z= t+ iy0 where y0 is the imaginary part of z0 and t can be negative real number.