Branches of log [closed]

I don't really understand what branches of log are in complex analysis. The definition I have is that 'a branch of log $$z$$ in G is a continuous function $$l$$ in G such that, for each $$z$$ in G, the value $$l(z)$$ is a logarithm of z.

I know this definition, but somehow still have a hard time understanding the concept. I may be overthinking it... I thought a branch of log z was essentially an identical log of z, just rotated around the plane by a constant times 2π. So just the same log z, but on top or underneath the original one (if you graphed it using $$\theta$$ as the z axis). Is this correct? Does this mean that there is only one branch in each [0,2π] interval? (or is this interval open?)

closed as unclear what you're asking by A. Goodier, Lord Shark the Unknown, Leucippus, Lee David Chung Lin, CesareoMay 9 at 11:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What don't you understand from the Wikipedia page on this subject? – David G. Stork May 8 at 20:25
• It is equivalent to defining the polar coordinate $\theta$ in $z = r\exp(i\theta)$. The most general definition involves an arbitrary non-self intersecting oriented curve that starts at the origin and moves away to infinity that defines the branch cut across which $\theta$ will jump by $2\pi$. – Count Iblis May 8 at 20:33
• Do you understand $\arcsin$ or $\arccos$? – qbert May 8 at 21:29
• The common branches are for $t \in (a,a+1), r \in (0,\infty), L_a(r e^{2i\pi t}) = r+2i\pi t$. But there are many more : for $t \in (0,\infty)$ and $r \in (2^t,2^{t+1})$ let $\ell(r e^{2i \pi t}) = r+2i\pi t$, this is a weird branch of $\log$ whose domain of analyticity and continuity is a spiral. – reuns May 8 at 22:50

We seek an inverse $$f(z)$$ to the exponential function. So we need $$f \circ \exp(z)=z$$. Since $$\exp{z}= \exp{(z+2i\pi)}$$ this identity cannot be simultaneously satisfied by $$z$$ and $$z'$$ if $$z-z'$$ is a multiple of $$2\pi i$$. Hence we have to assume that the domain contains only one representative of each equivalence class $$z+2i\pi \mathbb{Z}$$.