Complex logarithm

For $z=r{e}^{i\theta }\ne 0$, the complex logarithm has the values $\log r+i(\theta +2\pi k)$ for $k\in \mathbb{Z}$.

A logarithmic singularity behaves locally like $a\log (z-z_0)$ with $a\ne 0$, on a chosen branch. Although unbounded at the endpoint, it is integrable along a finite ray. In a rapidly oscillating endpoint integral its contour rotation produces a logarithm of the large parameter and a branch-angle contribution; this differs from the ordinary smooth-amplitude Watson lemma.

A branch of the complex logarithm chooses the argument continuously on a domain avoiding an appropriate branch cut, making the logarithm single-valued there.

The principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$.

If $\ell$ is an analytic branch of the logarithm on a connected domain avoiding zero, then $\exp (\ell /2)$ is an analytic branch of the square root. Any two such branches differ everywhere by one constant sign.

For $z=r{e}^{i\theta }$ with $-\pi <\theta \le \pi$, the principal square root is $\sqrt{r}{e}^{i\theta /2}$. Its standard analytic domain removes the nonpositive real axis.

For nonzero complex $z$ and complex $w$, the multivalued power is

The principal cube root is $z^{1/3}=\exp (\mathrm{Log}z/3)$ using the principal complex logarithm. In the right half-plane, its complex argument lies between $-\pi /6$ and $\pi /6$. Consequently the roots of $r^3=-z$ consist of one root with negative real part, $-z^{1/3}$, and two with positive real part, $z^{1/3}{e}^{\pm i\pi /3}$.

For every integer $n$ and real $\theta$,It converts powers of a complex number in polar form into multiplication of its complex argument.

On $\mathbb{C}\setminus [0,\infty )$, choose the argument in $(0,2\pi )$. Thenis an analytic logarithm. Locally this follows by scaling the convergent series

For $-1<\alpha <1$,with value one at $\alpha =0$ by continuity. A keyhole contour around the positive axis gives a jump factor $1-e^{2\pi i\alpha }$, while the double pole at $-1$ has residue $-\alpha e^{\pi i\alpha }$.