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An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

An argument of a nonzero complex number $z$ is any angle $\theta$ for which $z=|z|e^{i\theta }$. All arguments are $\theta +2\pi k$, with $k$ an integer. Zero has no argument: it has no distinguished direction. A commonly chosen representative, the principal argument $\mathrm{Arg}z$, lies in $(-\pi ,\pi ]$. Choosing representatives is part of specifying a branch of a multivalued function. These conventions are described in Jeremy Orloff, MIT 18.04, topic 1, sections 1.5 and 1.14.

For $3+4i$, the principal argument is $\arctan (4/3)\approx 0.927295$ radians, computed with the inverse tangent. But the coordinate ratio alone loses the quadrant: $-3-4i$ has the same ratio $4/3$, yet its principal argument is $\arctan (4/3)-\pi \approx -2.214297$ radians.

For nonzero $z,w$, the correct rule isOrdinary equality can fail. With $z=w=e^{3\pi i/4}$, their principal arguments sum to $3\pi /2$, but $zw=-i$ has principal argument $-\pi /2$. They differ by one full turn. As a point crosses the negative horizontal axis, the principal argument jumps by $2\pi$. A continuous choice on a suitable region therefore requires a branch cut; the jump belongs to the choice of representative, although the point moves smoothly.

A branch of a multivalued analytic expression is a single-valued holomorphic function on a specified domain that consistently selects one of the expression's possible values. A branch cut is a convenient way to choose such a domain, rather than an intrinsic locus of singularities. On $D=\mathbb{C}\setminus (-\infty ,0]$, the principal logarithm supplies two square-root branchesFor $z=i$, these give $B_{+}(i)=e^{i\pi /4}=(1+i)/\sqrt{2}$ and $B_{-}(i)=-(1+i)/\sqrt{2}$. Their derivatives follow by differentiating $B_{\pm }^2=z$: $B_{\pm }^{\prime }=1/(2B_{\pm })$, so $B_{+}^{\prime }(i)=(1-i)/(2\sqrt{2})$. There are exactly two branches on $D$: for any other holomorphic square root $B$, the ratio $B/B_{+}$ is continuous, takes values in $\{1,-1\}$, and hence is constant on the connected domain.

The cut can move. Choosing $0<\theta <2\pi$ defines a holomorphic logarithm $L_{+}(z)=\log |z|+i\theta$ on $D_{+}=\mathbb{C}\setminus [0,\infty )$. At $-1$ this gives $L_{+}(-1)=i\pi$ and $\exp (L_{+}(-1)/2)=i$, both regular values despite $-1$ lying on the previous cut. In the complex upper half-plane, $L_{+}$ agrees with $\mathrm{Log}$; in the lower half-plane it equals $\mathrm{Log}+2\pi i$, so the corresponding square roots have opposite signs there. The overlap $D\cap D_{+}$ has two components, exactly why the identity theorem cannot force agreement between them. ^[ref]

For example, the limits of $B_{+}$ at $-r$, $r>0$, are $i\sqrt{r}$ from above and $-i\sqrt{r}$ from below. This jump records the chosen branch cut; each point $-r\ne 0$ still has a neighbourhood supporting a holomorphic square root. Moving the cut changes the bookkeeping, while the branch point at $0$ and its monodromy persist. Specifying a branch of a multivalued function is therefore essential before differentiating or performing contour integration with holomorphic square roots, complex logarithms or complex exponentiation.

A holomorphic logarithm of a nowhere-zero holomorphic function $f$ on a domain $U$ is a holomorphic function $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the chain rule gives $g^{\prime }=f^{\prime }/f$. For $f(z)=z$, a holomorphic logarithm is a branch of the complex logarithm, and its derivative is $1/z$.

The principal logarithm is holomorphic on the slit complex plane $D=\mathbb{C}\setminus (-\infty ,0]$, withHere $\mathrm{Arg}$ is the chosen complex argument, and $\mathrm{Im}\mathrm{Log}z\in (-\pi ,\pi )$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this holomorphic domain. ^[ref] For example, $1+i=\sqrt{2}{e}^{i\pi /4}$ gives $\mathrm{Log}(1+i)=\frac{1}{2}\log 2+i\pi /4$ and $(\mathrm{Log}{)}^{\prime }(1+i)=(1-i)/2$; all its other complex logarithms differ by $2\pi ik$, $k\in \mathbb{Z}$.

There is no holomorphic logarithm of $z$ on the punctured complex plane $\mathbb{C}\setminus \{0\}$. Such a function would be a holomorphic primitive of $1/z$, so its contour integral around every closed path would vanish. On the counterclockwise unit circle, however,This is the period obstruction to a holomorphic antiderivative, and also exactly the logarithm change in the monodromy diagram.

Every nowhere-zero holomorphic function $f$ on a simply connected domain $U$ does admit a holomorphic logarithm. Choose $a\in U$ and $b\in \mathbb{C}$ with $e^b=f(a)$, and defineThe Cauchy integral theorem makes this contour integral independent of the path in $U$. Then $g^{\prime }=f^{\prime }/f$ and $(f{e}^{-g}{)}^{\prime }=e^{-g}(f^{\prime }-f{g}^{\prime })=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a connected domain differ by a constant $2\pi ik$, since their difference is continuous and takes values in the discrete set $2\pi i\mathbb{Z}$. ^[ref]

For a computed example, take $f(z)=1+z$ on the unit disc, with $a=b=0$. Expanding $f^{\prime }/f=1/(1+z)$ as a geometric series and integrating givesThe construction also produces holomorphic square roots: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent branch of a multivalued function for complex exponentiation by $\alpha$. A holomorphic logarithm converts multiplicative expressions into additive ones while making the domain restrictions explicit.

The inverse of the tangent restricted to $(-\pi /2,\pi /2)$, taking real values in that interval. Its derivative is $(1+x^2{)}^{-1}$. A formula involving an inverse tangent of a ratio may need a continuous branch of a multivalued function to describe a smooth field through a pole of that ratio.

Monodromy records how a germ of a holomorphic function changes after analytic continuation around a closed path. The source point returns to itself, but the continued germ of a holomorphic function need not. For $z(t)=e^{it}$, $0\le t\le 2\pi$, starting with holomorphic square root value $1$ gives $w(t)=e^{it/2}$ and hence $w(2\pi )=-1$; starting with complex logarithm value $0$ gives $L(t)=it$ and hence $L(2\pi )=2\pi i$. After $m$ counterclockwise turns the values are $(-1{)}^m$ and $2\pi im$. Thus monodromy exchanges the two square-root branches of a multivalued function, whereas it shifts through infinitely many logarithm branches of a multivalued function. ^[ref]

The three panels use the same parameter $t$. They show analytic continuation through local holomorphic patches, including across a chosen branch cut; repeatedly evaluating the principal logarithm would instead reset its imaginary part at the cut. Monodromy makes the obstruction to a global branch of a multivalued function visible and motivates the use of Riemann surfaces as domains on which continued values stay distinct.

Choose the principal branch of a multivalued function, $-\pi <\arg s<\pi$, for $s^{1/2}$. The integrand $G(s)=e^{st}/((s^2+1)\sqrt{s})$ has simple poles at $\pm i$ and a branch cut along the negative real axis. Its residues arewhose sum is $\sin (t-\pi /4)$.

For $t>0$, close the Bromwich contour to the left using a rectangle with outer real part $-R$ and horizontal edges at imaginary parts $\pm R$, indenting the cut and zero. On every distant edge, $|s^2+1|\ge |s{|}^2-1$, $|\sqrt{s}|=|s{|}^{1/2}$ and $|e^{st}|\le e^{\gamma t}$. Each edge has length $O(R)$ and $|s|\ge R$, so its integral is $O(R^{-3/2})$. The small circle of radius $\epsilon$ about zero contributes $O({\epsilon }^{1/2})$. The tails of the original vertical integral are absolutely bounded by $O(R^{-3/2})$ as well.

On the upper and lower banks at $s=-r$, the square roots are $i\sqrt{r}$ and $-i\sqrt{r}$, respectively. With the upper bank traversed toward zero and the lower bank away from zero, their combined integral isThe cut integral converges near zero since $r^{-1/2}$ is integrable, and at infinity by exponential decay. The residue theorem, followed by $R\to \infty$ and $\epsilon \to 0$, therefore givesFor $t<0$, close the Bromwich contour to the right. There are no enclosed poles or cuts, and $|e^{st}|\le e^{\gamma t}$ there. The same distant-edge estimates show $f(t)=0$ for $t<0$.

For the large-$t$ expansion, $1/(1+r^2)=1-r^2+r^4/(1+r^2)$. Scaling $r=u/t$ and using the given integral yields ${\int }_0^{\infty }{e}^{-rt}{r}^{-1/2}dr=\sqrt{\pi }{t}^{-1/2}$. Integration by parts twice gives the next coefficient $\Gamma (5/2)=3\sqrt{\pi }/4$. The remainder is bounded by ${\int }_0^{\infty }{e}^{-rt}{r}^{7/2}dr=\Gamma (9/2)t^{-9/2}$. Thus, without needing a globally convergent termwise integration of the local power series,In particular, the requested two leading terms are $\sin (t-\pi /4)+(\pi t{)}^{-1/2}$, interpreted as an additive asymptotic expansion. This is Bromwich inversion with a square-root branch cut.

For the all-kink sector, keep ${\alpha }_{ij}={\tanh }^2[({\theta }_i-{\theta }_j)/2]$ and $E_i=e^{X_i}$. Enumerating the even and odd binary configurations gives the Hirota tau functionsThe last minus sign is the product of three negative pair coefficients. As in the two-body limit, the three velocities are $v_i={\beta }_i/{\kappa }_i=\tanh {\theta }_i$.

To follow the first kink, keep $X_1$ bounded and take $t\to \pm \infty$. Let $J_{\pm }$ be the set of spectators whose exponential diverges in that limit:For no large spectators, $g/f\to E_1$. For one large spectator $j$, $g/f\to -1/({\alpha }_{1j}{E}_1)$. For two large spectators, the dominant terms give $g/f\to {\alpha }_{12}{\alpha }_{13}{E}_1$. On a continuous branch of a multivalued function, each case has local profile $4\arctan \exp (X_1+c_{\pm })$ plus the appropriate vacuum offset, whereConsequently the incoming and outgoing intercepts are $-({\gamma }_1+c_{-})/{\kappa }_1$ and $-({\gamma }_1+c_{+})/{\kappa }_1$. This provesAgain the time expression requires $v_1\ne 0$ and distinct velocities; the corresponding spatial-shift identity holds also when $v_1=0$. With mixed orientations, use the general pair shift established above and determine growing spectators by the sign of ${\kappa }_j(v_1-v_j)$; the same multiplication of pair coefficients proves additivity.

There is no independent three-body contribution to the asymptotic shift. The pairwise additivity of soliton shifts is a classical manifestation of factorized scattering in an integrable partial differential equation. The collision preserves the individual asymptotic rapidities and profiles, and the net shift is independent of the sequence of separated pair collisions. In the quantum theory, consistency of the corresponding species-changing S-matrices becomes the Yang-Baxter equation; the classical scalar shift identity is its physical precursor, rather than a derivation of all quantum matrix identities.

Write the rapidity parameters as ${\kappa }_i=\cosh {\theta }_i$, ${\beta }_i=\sinh {\theta }_i$, and define $E_i=e^{X_i}$. The signed coefficient in the Sine-Gordon multisoliton tau representation isFor distinct rapidities, $0<{\alpha }_{12}<1$. In particular, $B_{12}$ cannot be taken as a real logarithm of a positive coefficient. The finite sums defining the Hirota tau functions can instead be evaluated directly with the real, negative $a_{12}$. They giveThe physical field is a continuous branch of a multivalued function, equivalently $\varphi =4\arg (f+ig)$ with the argument followed continuously. The principal inverse tangent alone jumps when $f$ changes sign.

Follow the first kink with $x=v_{1}t+O(1)$. Then $X_1=O(1)$ and $X_2={\kappa }_2(v_1-v_2)t+O(1)$. The two possible local limits arewhere the second field is written on the continuous kink branch. Thus both limits are single Sine-Gordon kinks of the same width and velocity, but their centers obey $X_1=0$ or $X_1+\log {\alpha }_{12}=0$. Following the second kink gives the same conclusion with labels exchanged. The incoming and outgoing velocities are thereforeThere is no change in the asymptotic rapidities or kink profiles.

Define the spatial shift as the outgoing center intercept minus the incoming center intercept. Since the large-$E_2$ limit occurs afterwards when $v_1>v_2$, and beforehand when $v_1<v_2$, the soliton time delay isThe time formula uses $\Delta t=-\Delta x/v_1$ and requires $v_1\ne 0$. Its dependence on the velocities is explicit on substitutingFor a faster right-moving kink, $\Delta x_{1|2}>0$ and $\Delta t<0$: it arrives earlier than its freely continued incoming trajectory. If $v_1=0$, report the finite spatial shift; a fixed-position arrival-time delay for a stationary kink is undefined. Coincident velocities are excluded from a separated collision asymptotic.

For completeness, allowing antikinks means ${\kappa }_i={\sigma }_i\cosh {\theta }_i$, ${\beta }_i={\sigma }_i\sinh {\theta }_i$, with ${\sigma }_i=\pm 1$. The velocities remain $\tanh {\theta }_i$. For opposite orientations, $|a_{12}|={\coth }^2[({\theta }_1-{\theta }_2)/2]$, and the general spatial shift isThis follows from the same two local limits; it makes explicit the orientation hypothesis behind the velocity-only all-kink answer.

On $\mathbb{C}\setminus [-1,1]$, the chosen branch of a multivalued function is characterized by $f(z{)}^2=z^2-1$ and $f(x)>0$ for $x>1$; equivalently $f(z)/z\to 1$ as $z\to \infty$. Define $h(z)=-f(-z)$. Then $h(z{)}^2=z^2-1$ and $h(z)/z\to 1$ at infinity, so uniqueness of the normalized holomorphic square root gives $h=f$. Therefore $f(-z)=-f(z)$.

Harmonic parts. Write the holomorphic function as $f=u+iv$. The Cauchy-Riemann equations say $u_x=v_y$ and $u_y=-v_x$. Equality of mixed partial derivatives therefore givesThus both the real and imaginary parts are harmonic functions and satisfy two-dimensional Laplace's equation.

Single-valued logarithm. A convenient choice isusing the principal logarithm for each term. Its branch cuts in the $z$ plane are $(-\infty ,-1]$ and $[1,\infty )$. The branch points are $-1$ and $1$: going once around the first adds $2\pi i$, while going once around the second subtracts $2\pi i$. Removing the two exterior rays makes this a consistent branch of a multivalued function and in particular leaves the whole open unit disk available.

Potential inside the cylinder. Put $z=r{e}^{i\theta }$ with $r<1$. Multiplication by $1-\overline{z}$ givesThe denominator is positive and the real part of the numerator is positive. Thus its argument lies strictly between $-\pi /2$ and $\pi /2$, andAs $r\uparrow 1$, this tends to $\pi /2$ on the upper open semicircle and $-\pi /2$ on the lower open semicircle. Scale the imaginary part to obtainIt is a harmonic function, has the required boundary values, and gives $\Phi (0)=0$. The two joining points of the boundary arcs have discontinuous boundary data, so no common continuous boundary value is demanded there. Along the real diameter, including radial approaches to either joining point, the displayed solution is zero. This is the disk potential with two semicircular boundary values.

For a second check, the Taylor series of the logarithms gives $f(z)=2{\sum }_{j\ge 0}{z}^{2j+1}/(2j+1)$. Taking the imaginary part yields the Fourier serieswhose boundary limit is exactly the required alternating pair of constant semicircle values.

Poles and residues. An isolated singularity $z_0$ is a pole of order $m$ if its Laurent series has finitely many negative powers, the lowest being a nonzero multiple of $(z-z_0{)}^{-m}$. The residue is the coefficient of $(z-z_0{)}^{-1}$.

If $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, factor $h(z)=(z-z_0)H(z)$, where $H$ is holomorphic near $z_0$ and $H(z_0)=h^{\prime }(z_0)$. For $g(z_0)\ne 0$,The last quotient is holomorphic and nonzero at $z_0$, proving a simple pole with

The given function. For noninteger $a$, specify a branch of a multivalued function before using $z^a$. Choose $z^a=\exp (a\log z)$ with $-\pi /2<\arg z<3\pi /2$, a cut along the negative imaginary axis. This leaves all three roots of $1+z^3$ off the cut. They are $z_{+}=e^{i\pi /3}$, $z_{-}=e^{-i\pi /3}$ and $z_0=-1$, all simple. The numerator is nonzero at these points. Since the denominator has derivative $3z^2$, the residues areFor noninteger $a$, zero is a branch point, rather than an isolated pole. This remains true when $-1<a<0$ and the numerator diverges there. For nonnegative integer $a$ it is a regular point. Other logarithm branches change the phases of the residues consistently.

Integral and convergence. Near zero the integrand behaves as $x^a$, requiring $a>-1$. At infinity it behaves as $x^{a-3}$, requiring $a<2$. Therefore the exact convergence range isUse a counterclockwise sector contour with rays at angles $0$ and $2\pi /3$, inner radius $\epsilon$ and outer radius $R$. The same complex logarithm branch is valid throughout this sector. Only $z_{+}=e^{i\pi /3}$ lies inside. The small circular arc contributes $O({\epsilon }^{a+1})$ and the large one contributes $O(R^{a-2})$, so both vanish precisely in the stated range.

The outward positive-real ray contributes $I$. On the returning ray set $z=x{e}^{2\pi i/3}$: the denominator is still $1+x^3$, while $z^{a}dz=e^{2\pi i(a+1)/3}{x}^{a}dx$, with reversed limits. The residue theorem givesPut $\alpha =\pi (a+1)/3$. The right-hand phase is $-e^{i\alpha }$ and $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$. HenceThe sine is positive in this range, consistent with a positive real integral. This is a special case of the Mellin integral of a reciprocal binomial.

Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.

Choose a continuous branch of a multivalued function of the complex logarithm on a simply connected region avoiding zero. ThenAlternatively $u_r=-e^{-\theta }\sin (\ln r)/r=v_{\theta }/r$ and $v_r=e^{-\theta }\cos (\ln r)/r=-u_{\theta }/r$, verifying the Cauchy-Riemann equations. The holomorphy is local to a chosen branch of a multivalued function: increasing $\theta$ by $2\pi$ multiplies $g$ by $e^{-2\pi }$, so this is not a single-valued holomorphic function on the whole punctured plane.

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

Shrink the positive outer contour to the branch cut $[-1,1]$. On its upper bank the chosen branch of a multivalued function has $\sqrt{x^2-1}=i\sqrt{1-x^2}$ and on the lower bank it has the negative of this value. The upper bank runs from $1$ to $-1$, and the lower bank runs from $-1$ to $1$. The endpoint circles contribute $O(\sqrt{ϵ})$ and vanish. HenceThe change of variables $x=\sin t$ independently reduces the result to the integral of ${\sin }^{4}t$ over $[-\pi /2,\pi /2]$.