To do calculus with the logarithm we need a function, not a set of values, and we need it to be continuous. A branch picks one value of $\log z$ at each point in a way that varies continuously. Every branch must leave out a curve from $0$ to $\infty$ (a branch cut), because going once around $0$ changes any continuous choice of argument by $2\pi$. The cut is a region boundary that we choose.
Key idea
$\operatorname{Log} z$ is continuous, and analytic with $(\operatorname{Log} z)' = \frac{1}{z}$, on the plane with the ray $(-\infty, 0]$ removed. Across that cut it jumps by $2\pi i$: at $-a$ ($a > 0$) the values from above tend to $\ln a + \pi i$ and from below to $\ln a - \pi i$.
Choosing arguments in $(\alpha, \alpha + 2\pi]$ instead gives another branch, with its cut along the ray at angle $\alpha$. Likewise, $\operatorname{Log}(g(z))$ is analytic except where $g(z)$ is a real number $\le 0$.
Why it works
$\operatorname{Arg} z$ is continuous everywhere except on the negative real axis, where it jumps from near $\pi$ to near $-\pi$; $\ln|z|$ is continuous for $z \ne 0$. So $\operatorname{Log}$ is continuous off the cut. There it is the inverse of $e^w$ on the strip $-\pi < \operatorname{Im} w < \pi$, and the inverse function rule gives $\frac{d}{dz}\operatorname{Log} z = \frac{1}{e^w} = \frac{1}{z}$. No continuous branch can live on a whole circle around $0$: following it once around would change the argument by $2\pi$, so the value could not match up when you return.
Worked examples
Worked example: Approaching the cut
Find the limits of $\operatorname{Log} z$ as $z \to -4$ from above and from below.
- $\ln|z| \to \ln 4$ either way.
- From above, $\operatorname{Arg} z \to \pi$; from below, $\operatorname{Arg} z \to -\pi$.
- The limits are $\ln 4 + \pi i$ and $\ln 4 - \pi i$: a jump of $2\pi i$.
Answer: $\ln 4 \pm \pi i$
Worked example: Another branch
The branch $F$ of $\log z$ with arguments in $(0, 2\pi]$ has its cut on the positive real axis. Find $F(-1)$ and $F(-i)$.
- $\arg(-1)$ in $(0, 2\pi]$ is $\pi$: $F(-1) = \pi i$ (the same as $\operatorname{Log}$).
- $\operatorname{Arg}(-i) = -\frac{\pi}{2}$, which is not in $(0, 2\pi]$; adding $2\pi$ gives $\frac{3\pi}{2}$.
- $F(-i) = \ln 1 + \frac{3\pi}{2}i = \frac{3\pi}{2}i$, while $\operatorname{Log}(-i) = -\frac{\pi}{2}i$.
Answer: $\pi i$ and $\frac{3\pi}{2}i$
Worked example: Where is a composition analytic?
Where is $\operatorname{Log}(2z + 6)$ analytic?
- $\operatorname{Log} w$ fails exactly when $w$ is real and $w \le 0$.
- $2z + 6$ is real and $\le 0$ exactly when $z$ is real and $z \le -3$.
- So the function is analytic on the plane minus the ray $(-\infty, -3]$, with derivative $\frac{2}{2z + 6} = \frac{1}{z + 3}$.
Answer: everywhere except $(-\infty, -3]$
Common mistake
Two opposite errors. (1) Treating $\operatorname{Log} z$ as analytic everywhere except $0$: it is not even continuous on the negative real axis, so, for example, it cannot serve as an antiderivative of $\frac{1}{z}$ along a loop around $0$. (2) Thinking the cut belongs to the logarithm itself: the many-valued $\log z$ has no cut; a cut is a choice we make to get a single-valued branch, and a different choice gives a different branch. Check by asking: on which set is the formula I wrote continuous?