Analytic continuation

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.

A boundary of a domain of a holomorphic function is natural if the function has no analytic continuation through any of its points. For $f(z)={\sum }_{n\ge 1}{z}^{2^n}$, each root of unity of order a power of two is a singular boundary point: along $z=r\zeta$ all sufficiently late terms equal $r^{2^n}$ and their sum tends to infinity as $r\uparrow 1$. These roots are dense on the unit circle, so no boundary arc admits analytic continuation.

Zeta function regularization assigns a finite value to a divergent spectral sum by first forming a convergent complex-power series and then using its analytic continuation. For positive frequencies ${\omega }_n$, continue ${\sum }_n{\omega }_n^{-s}$ to $s=-1$ to define a regulated frequency sum, when that continuation is regular there. This does not turn the divergent positive-term sum into an ordinary convergent sum. It is a useful subtraction convention for the normal-ordering constant of a string and for vacuum energies.

For $r=n+1/2$, $n\ge 0$, the convergent frequency zeta function, expressed through the Riemann zeta function, is ${\sum }_r{r}^{-s}=(2^s-1){\zeta }_R(s)$ when $\mathrm{Re}s>1$. Its analytic continuation at $s=-1$ gives $(1/2-1)(-1/12)=1/24$. A common exponential frequency cutoff gives ${\sum }_{r}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, so the same finite part follows by subtracting the leading divergence. In contrast, integer frequencies have regulated sum $-1/12$.

A meromorphic continuation extends a holomorphic function from a connected open set to a larger connected domain as a meromorphic function, allowing isolated poles. Equality on the initial open set determines the extension uniquely by the identity theorem. Integral formulas whose integrands depend holomorphically on a parameter often supply such an extension after contour deformation or subtraction of singular terms.

As distributions on the real line,The Sokhotski--Plemelj formula relates the two boundary values of a Cauchy transform across its contour and fixes the spectral delta functions generated by retarded and advanced prescriptions.

On a positively oriented smooth simple closed contour, set $\mathcal{C}f=(\pi i{)}^{-1}\mathrm{PV}{\int }_{\Gamma }f(t)/(t-z)dt$. The Sokhotski–Plemelj theorem gives the sum and difference of the two Cauchy boundary values. Changing the sign of the exterior branch converts the equation $\mathcal{C}f=g$ into a normalized additive Riemann-Hilbert problem with jump $g$. Applying the boundary formula again yields $f=\mathcal{C}g$, proving the involution on a suitable smooth or Hölder function class.

For $0<\alpha <1$ and $-1<x<1$, the endpoint powers are integrable and the interior pole is taken as a Cauchy principal value. The analytic function $w(z)=(z-1{)}^{\alpha -1}/(z+1{)}^{\alpha }$, positive for $z>1$ and decaying like $1/z$, has boundary values $-e^{\pm i\pi \alpha }(1-t{)}^{\alpha -1}/(1+t{)}^{\alpha }$. Its jump gives a Cauchy reconstruction, while the mean of its boundary values and the Sokhotski–Plemelj formula give the displayed identity. At $\alpha =1/2$ the principal value is zero.

Suppose $w$ is holomorphic off $[0,1]$, tends to zero at infinity, has sufficiently regular boundary values in the interval interior, and has endpoint growth $O(|z-a{|}^{-\alpha })$ for some $\alpha <1$ at each endpoint $a$. If its jump is integrable, the displayed reconstruction follows by deforming a Cauchy integral formula contour onto the two sides of the interval. An endpoint circle of radius $\epsilon$ contributes $O({\epsilon }^{1-\alpha })$, which vanishes. The two oppositely oriented banks leave precisely the jump. Without decay at infinity, an entire remainder can survive; a nonzero constant is the simplest example.

With $Hv(\rho )={\pi }^{-1}\mathrm{PV}\int v(s)/(\rho -s)ds$, the upper and lower limiting Cauchy integrals are $P^{\pm }v=\pm v/2+iHv/2$. They obey $P^{+}-P^{-}=I$. The actual complementary idempotent projections are $P^{+}$ and $-P^{-}$; confusing this signed convention with two positive projectors changes reconstruction signs. These boundary operators naturally enter the spectral reconstruction of an attenuated Radon transform.

The monodromy theorem says that, when analytic continuation exists throughout a domain, its ambiguity depends on homotopy. Let $D\subset \mathbb{C}$ be a domain, $a\in D$, and $g_a$ a germ of a holomorphic function at $a$. Assume $g_a$ admits analytic continuation along every continuous path in $D$ starting at $a$. Two such paths ending at $b$ give the same terminal germ of a holomorphic function if they are homotopic with endpoints fixed in $D$. ^[ref]

Here a homotopy with endpoints fixed is a continuous map $H:[0,1{]}^2\to D$ satisfyingEvery intermediate path must stay in $D$. Equality means agreement on a neighbourhood of $b$, rather than merely equal values at $b$. ^[ref]

The identity theorem provides local rigidity; the monodromy theorem shows precisely how topology controls the remaining global freedom. Deforming a route without crossing a forbidden point cannot change the resulting germ of a holomorphic function. Thus a complicated analytic continuation calculation can be replaced by one on a simpler homotopic route. Distinct homotopy classes can still give the same continuation: a rational function on its domain already gives a single-valued answer along every route, even around a pole.

On a simply connected domain $D$, a germ of a holomorphic function admitting analytic continuation along every path from its base point determines a unique holomorphic function $F:D\to \mathbb{C}$ with that initial germ of a holomorphic function. ^[ref]

Indeed, any two paths with the same endpoints are homotopic with endpoints fixed. Define $F(b)$ using the terminal germ of a holomorphic function. The monodromy theorem makes this independent of the path. In a small open disc about $b$, extend the chosen route by a short segment; the same local representative supplies $F$ throughout the open disc, proving holomorphy. The identity theorem proves uniqueness. The existence hypothesis of the monodromy theorem must still be checked.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

The monodromy theorem gives equality of continuations that exist under its hypothesis. It does not assert that every holomorphic function admits analytic continuation everywhere, even within a simply connected domain.

For an explicit obstruction, take $f(z)=1/(1-z)$ near $a=0$ and attempt to apply the theorem on $D=\mathbb{C}$. The path $\eta (t)=t$ ends at $1$, where no terminal germ of a holomorphic function exists. If a holomorphic representative $G$ near $1$ continued $f$, then the identity theorem would extend $(1-z)G(z)=1$ to that neighbourhood. Substituting $z=1$ would give $0=1$. Thus the pole prevents the existence hypothesis, although $\mathbb{C}$ is a simply connected domain. The analytic continuation article explains why the same rational function does continue throughout $\mathbb{C}\setminus \{1\}$.

Existence at an endpoint by some route is also weaker than existence along every path from the base point. Along the real segment from $1$ towards $0$, a holomorphic square root of $z$ tends to $0$, yet cannot continue holomorphically through $0$: a local representative $G$ would satisfy $G^2=z$ by the identity theorem, and differentiating at $0$ would give $2G(0)G^{\prime }(0)=1$ with $G(0)=0$. A finite limiting value therefore does not remove a branch point. More severe obstructions appear at a natural boundary of a holomorphic function.

Along one path, analytic continuation is a chain of function elements on overlapping open discs, each agreeing with the preceding germ of a holomorphic function near the crossing point. The identity theorem makes this chain unique along that fixed path. Equality at a crossing means equality on a neighbourhood, not just equality of a single value.

For the deformation step, fix one intermediate path $t\mapsto H(s_0,t)$ and its continuation. By compactness of $[0,1]$, choose a finite chain of representative open discs and subdivide the parameter so that each subarc fits in its assigned open disc, with consecutive junctions in overlap neighbourhoods where the representatives agree. Joint continuity of $H$ ensures that every sufficiently nearby intermediate path uses that same chain and those same overlaps. The endpoints stay fixed, so the terminal germ of a holomorphic function stays unchanged. Existence along every intermediate path supplies this argument at each $s_0$; local constancy and connectedness of $[0,1]$ give equality at $s=0$ and $s=1$. This proves the monodromy theorem without assuming that every analytic germ projection is a covering map.

For a covering map $p:X\to Y$ and base point $y$, the path lifting theorem lifts each based loop from each point of the fibre $p^{-1}(y)$. Its endpoint defines a permutation of that fibre; path reversal gives the inverse. The homotopy lifting theorem for a covering map makes the permutation depend only on the based homotopy class. These permutations form the monodromy group, the image of the fundamental group actionThe image is independent of the convention for composing routes; one may reverse the loop classes to obtain a left group action when route multiplication means following the first route and then the second.

For $p(w)=w^2$ on the punctured plane, the fibre over $1$ is $\{1,-1\}$, and a once-around based loop exchanges its two points. The monodromy group has two elements: only the parity of the winding number matters. For the covering map $p(w)=e^w$ from the complex plane to the punctured plane, the fibre over $1$ is $2\pi i\mathbb{Z}$, and a positive turn acts by $2\pi ik\mapsto 2\pi i(k+1)$. Its monodromy group is an infinite cyclic group. These are the geometric versions of logarithm and square-root continuation along different paths: the base point returns, while its lifted endpoint can move to another sheet.

A connected three-sheeted cover has transitive monodromy. If it also has a simple branch point, its monodromy contains a transposition and must be the full symmetric group on three letters. A local triple branch supplies a three-cycle explicitly.

The degree-four rational map $F(z)=z^2+z^{-2}$ has branch values $-2,2,\infty$. Away from their inverse images, the transformationsare commuting deck involutions. They generate a Klein four-group acting transitively on each fibreConsequently the monodromy group is $V_4$, acting by the identity and the three double transpositions.

If a parameter-dependent contour integral has a moving pole, deforming the integration contour continuously so that the pole never crosses it preserves a holomorphic branch. Returning the contour to its original path after crossing a pole adds or subtracts the corresponding residue contribution.

For a figure-eight contour around $1$ anticlockwise and $-1$ clockwise, start $(t^2-1{)}^z$ at $t=0$ with argument $-\pi$. Shrinking the two loops gives the displayed identity. If the interval integral instead uses $(t^2-1{)}^z=e^{-i\pi z}(1-t^2{)}^z$, the corresponding identity includes the extra factor $e^{i\pi z}$. Omitting this phase changes the value, rather than merely the locations of the meromorphic continuation's poles.

The solutions of $\sinh w=z$ arefor any chosen local branch $\mathrm{Arcsinh}$.

It is the complex multivalued sine member of the inverse hyperbolic functions.