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 continuous choice of complex logarithm on a negative real interval has a fixed integer $k$ in the displayed expression. The choice with argument $\pi$ is the upper-bank boundary value of the principal logarithm; the usual analytic principal-branch domain itself excludes the negative axis. Consequently, integrating a real weight times that logarithm adds $(2k+1)i\pi$ times the integral of the weight to the real logarithmic integral. The branch choice must be stated.

Take a complex logarithm whose radial branch cut has angle $\beta$. On a positively oriented circle, use the one-sided branch values on either side of its crossing with the cut. Parametrize from $\theta =\beta$ to $\beta +2\pi$; adding an integer multiple of $2\pi i$ to the branch does not change the integral because the integral of $dz$ is zero. Along the open arc, a primitive is $z\mathrm{Log}z-z$. Its endpoint difference is $2\pi i$ times the cut-crossing point $R{e}^{i\beta }$, proving the displayed formula. The circle crosses the cut, so this is a piecewise contour integral of boundary values, not the integral of a single holomorphic logarithm on a neighborhood of a closed circle. The Cauchy integral theorem therefore does not make it zero.

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 $-\pi <\mathrm{Arg}z\le \pi$.

The power series $\mathrm{Log}(1-w)=-{\sum }_{j\ge 1}{w}^j/j$ proves this estimate. There is no global bound $|\mathrm{Log}(1-w)|\le C|w|$, since the logarithm diverges as $w\to 1$. The different estimate $\log |1-w|\le \log (1+|w|)\le |w|$ is valid wherever $w\ne 1$, and is sufficient for growth bounds on canonical products.

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.

If an entire function $h$ satisfied $h(z{)}^2=z$, then $h(0)=0$. Differentiating would give $2h(0)h^{\prime }(0)=1$, a contradiction. More generally, the order of every zero of a square of a holomorphic function is even, so a simple zero prevents a holomorphic square root on any neighborhood containing it. A square root can still exist on a slit domain avoiding the zero. The identity theorem can force this obstruction from values specified only on a sequence accumulating elsewhere in a connected domain.

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.

This extends real exponentiation.

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

For $w\ne 0$, different complex logarithm values differ by $2\pi ik$. Applying the exponential function gives the displayed multivalued power. Selecting a single logarithm branch gives one value. When $z=i$, successive branch values differ by the positive factor $e^{-2\pi }$ and lie on one ray of fixed argument $\ln |w|$. If this factor is not one, the nonzero values accumulate at zero in one direction of the integer index.

For $z\ne 0$, every logarithmic value is $\log |z|+i(\arg z+2\pi k)$ with integer $k$. Multiplying by $z$ and taking the real part proves the displayed formula for the modulus of each value of complex exponentiation. For nonreal $z$, these values form a geometric progression with nonunit positive ratio $e^{-2\pi \mathrm{Im}z}$. For real nonzero $z$, changing the logarithm branch changes only the phase and leaves the modulus fixed. A fixed initial argument changes only the indexing of the set of values.

Using principal complex logarithms, the inner power is $i^i=e^{-\pi /2}$. Consequentlywhich are different. Thus parentheses in nested complex exponentiation cannot be dropped or moved. There is an additional issue when all complex logarithm branches are allowed: each nested operation must retain its own branch choices. In the first expression, the complete set of values is $-i{e}^{-2\pi m}$, $m\in \mathbb{Z}$; in the second, it is $\exp [i(\pi /2+2\pi m)e^{-(\pi /2+2\pi k)}]$, $k,m\in \mathbb{Z}$. The principal-value comparison already establishes nonassociativity; the multivalued sets make the role of branching explicit.

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 an entire function $f$, use the complex logarithm $L$ with argument in $(0,2\pi )$. The unit-circle integral in the title uses the branch's one-sided boundary values at its crossing with the cut. On a keyhole contour, the outward upper bank has $L=\log x$, and the inward lower bank has $L=\log x+2\pi i$. Their combined integral of $f(L-i\pi )$ is $-2\pi i{\int }_{\epsilon }^{1}f(x)dx$. The inner circle contributes $O(\epsilon |\log \epsilon |)$, which vanishes. The Cauchy integral theorem on the slit plane therefore gives the identity.

On $z=e^{i\theta }$, $|L-i\pi |=|\theta -\pi |\le \pi$. The estimation lemma consequently givesSubtracting the middle branch value $i\pi$ centres the weight and produces this bound. The contour crosses the cut only in the limiting boundary-value interpretation, so it is not a closed contour inside the logarithm's domain.

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 $-\alpha e^{\pi i\alpha }$.