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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.

The displayed supremum norm is finite on every entire function, positive on every nonzero one by the identity theorem, and submultiplicative under pointwise multiplication. It is not complete: partial geometric sums ${\sum }_{j=0}^n(z/2{)}^j$ are Cauchy on the disk but their limit there extends meromorphically with a pole at two. More generally, no complete algebra norm exists on all entire functions because the coordinate function has algebraic spectrum of an element equal to the whole complex plane.

If $f$ is holomorphic on an open set $U$, its conjugate reflection $f^{\ast }$ is holomorphic on $U^{\ast }=\{z:\overline{z}\in U\}$, with $(f^{\ast }{)}^{\prime }(z)=\overline{f^{\prime }(\overline{z})}$. The identity follows by conjugating the difference quotient with increment $\overline{h}$; the two conjugations restore complex differentiability.

If $U$ is invariant under complex conjugation and is connected, and $f$ is real at a set of real points accumulating inside $U$, then $f=f^{\ast }$ by the identity theorem. In particular, an entire function real at every $1/n$ has this symmetry. This is a symmetry statement on an already defined domain; Schwarz reflection also constructs extensions across a boundary.

The displayed formula is a submultiplicative norm on the algebra of entire functions: definiteness follows from the identity theorem. No submultiplicative norm on the whole algebra can be complete. The coordinate function $Z(z)=z$ has every complex number in its algebraic spectrum of an element, because $Z-\lambda$ vanishes somewhere and has no entire inverse. This contradicts compactness of spectra in a Banach algebra.

For an entire function $f(z)={\sum }_j{c}_j{z}^j$, equality $|c_n|={\max }_{|z|=r}|f(z)|/r^n$ at a fixed $r>0$ holds exactly when $f(z)=c{z}^n$. The coefficient is the average of $r^{-n}f(r{e}^{i\theta })e^{-in\theta }$. Equality in its modulus bound forces the continuous integrand to have constant modulus and argument, hence to be constant. The identity theorem then extends the monomial identity from the circle to the whole plane. The zero function and the constant case $n=0$ are included.

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.

A function element is a pair $(f,D)$ consisting of a nonempty disk and a holomorphic function on that disk. Its local value determines a germ of a holomorphic function. Elements agreeing on an overlapping neighborhood continue one another; chains of such continuations produce the complete analytic function generated by an element. An algebraic identity holding in one element persists along any continuation chain by the identity theorem.

Suppose the distinct complex numbers ${\beta }_k$ satisfy $|{\beta }_k|\le \delta <1$. The geometric sequences in the displayed formula belong to the absolutely summable sequence space and have dense linear span. Otherwise the Hahn-Banach theorem gives a nonzero annihilating functional, represented by a bounded sequence $b_j$. Its power series $F(z)={\sum }_{j\ge 1}{b}_j{z}^{j-1}$ is holomorphic in the unit disk and vanishes at every ${\beta }_k$. Compactness supplies an accumulation point inside that disk. The identity theorem forces $F=0$ and hence all its coefficients zero, a contradiction. This is a useful form of density by annihilators that converts a Banach-space problem into uniqueness of an analytic function.

A holomorphic function $F$ on a larger open set $V$ whose restriction to $U$ is the original holomorphic function $f$. A removable singularity is a point across which such an extension exists. On a connected larger domain, a holomorphic extension is unique by the identity theorem, provided $U$ is nonempty.

The curvature-minus-one hyperbolic metric on the punctured disk assigns distance $2\log n$ along the radial geodesic from $e^{-1}$ to $e^{-n^2}$. Its exponential distances therefore sum to a finite value; the triangle inequality gives the same conclusion for any fixed basepoint. Nevertheless no nonzero bounded holomorphic function has these zeros: the Riemann removable singularity theorem extends it at zero, and the identity theorem then makes it identically zero. Simple connectivity is necessary in the hyperbolic characterization of bounded analytic zero sets.

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.

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.

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.

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.

If an entire function $f$ is not a polynomial, then there is a point $z_0\in \mathbb{C}$ for which $f^{(n)}(z_0)\ne 0$ for every nonnegative integer $n$. Consequently every coefficient in the Taylor series of $f$ about $z_0$ is nonzero.

For each $n$, the function $f^{(n)}$ is not identically zero, since otherwise $f$ would be a polynomial. Its zero set is closed and has empty interior by the identity theorem, hence is a nowhere dense set. The Baire category theorem says that the complete metric space $\mathbb{C}$ cannot be the union of these countably many zero sets.

Set $S_n=\{x:f^{(n)}(x)=0\}$ and $E_n=\mathrm{int}{S}_n$. The sets $S_n$ are closed, $E_n\subset E_{n+1}$, and Baire's theorem on each interval shows that $\Omega ={\bigcup }_n{E}_n$ is dense. On every connected component of $\Omega$, compactness and the increasing cover by the $E_n$ show that one derivative vanishes locally with a uniform finite order, so $f$ agrees there with one polynomial.

If $F=\mathbb{R}\setminus \Omega$ were nonempty, it would be closed and have no isolated points: polynomial pieces on both sides of an isolated point would have matching derivatives of every order and hence join into one polynomial piece. Applying Baire's theorem to the cover $F={\bigcup }_n(F\cap S_n)$ gives an interval $U$ and an index $k$ for which $F\cap U$ is nonempty and contained in $S_k$. Because $F$ has no isolated points, difference quotients show that every derivative of order at least $k$ vanishes on $F\cap U$. Each polynomial component of $\Omega$ meeting $U$ has an endpoint in $F\cap U$, so its degree is less than $k$. Thus $f^{(k)}=0$ throughout $U$, contradicting $F\cap U\ne \varnothing$. Hence $\Omega =\mathbb{R}$, and its sole connected component carries one polynomial.

The derivative formula requires additional regularity: continuity alone does not define $f^{\prime }$. Assume $f$ is continuously differentiable, or locally absolutely continuous with its almost-everywhere derivative. Integration by parts on $[0,R]$ givesFor $\mathrm{Re}p>c$, the given exponential bound makes the first term tend to zero. Thus the improper transform exists under these regularity assumptions andFor the primitive, absolute integrability justifies reversing the integration order:Indeed the corresponding absolute double integral is finite when $\mathrm{Re}p>c$.

For real $p>0$, substitution $t=s^2/p$ and the supplied Gaussian integral giveBoth sides are holomorphic for $\mathrm{Re}p>0$, so the identity theorem extends this equality there, taking the square root with positive real part. The integrable singularity at zero causes no difficulty for this improper integral. The primitive rule then gives $\mathcal{L}[2\sqrt{t}]=\sqrt{\pi }{p}^{-3/2}$.

For the conductor, take the Laplace transform in time and write $U(x,p)$. The initial temperature gives $U_{xx}=pU$. Use the usual physical condition that the temperature is bounded far down the conductor; this selects the decaying solutionThe constant applied flux has transform $Q/p$, so $-k{U}_x(0,p)=Q/p$, giving $A(p)=Q/(k{p}^{3/2})$. This is the surface temperature under constant heat flux on a half-line. Using the transform just calculated yieldsHere $k>0$ is the thermal conductivity. The far-field boundedness is the customary physical selection condition; without a spatial growth restriction the written PDE and initial/boundary data alone do not specify that selection. Within the physical solution, the face temperature tends continuously to zero as $t\downarrow 0$.

At a zero $a$, Taylor's theorem gives $f(z)={\sum }_{n\ge 0}{c}_n(z-a{)}^n$ near $a$. If some coefficient is nonzero, let $m$ be the least such index. Then $f(z)=(z-a{)}^{m}q(z)$ with $q(a)\ne 0$, and continuity makes $q$ nonzero on a smaller disc. Thus $a$ is an isolated zero. A limit point of zeros is also a zero by continuity and cannot be isolated, so all Taylor coefficients there vanish and $f$ vanishes on a nonempty open disc.

Let $A$ be the set of points having a neighborhood on which $f$ vanishes. It is nonempty and open. A point in its relative closure either already belongs to $A$ or is a limit point of distinct zeros; the preceding argument puts it in $A$ in either case. Thus $A$ is also relatively closed. Connectedness of $D$ gives $A=D$, proving $f$ is identically zero.

For conjugate reflection, the disc is invariant under conjugation, andHence $h$ is holomorphic, with derivative $\overline{g^{\prime }(\bar{z})}$. If $g$ is real on the real diameter, then for real $x\in (-1,1)$, $h(x)=\overline{g(x)}=g(x)$. The holomorphic difference $g-h$ therefore has zeros accumulating at zero, and the identity theorem just proved gives $g=h$ on the entire disc.

For a holomorphic function on $|z-a|<R$, the complex Taylor theorem stateswith uniform convergence on each smaller closed disc. Here is a proof. Choose $r<\rho <R$. For $|z-a|\le r$ and $|\zeta -a|=\rho$, the geometric seriesconverges uniformly. Substitute it into Cauchy integral formula and integrate term by term to getIf $M_{\rho }={\max }_{|\zeta -a|=\rho }|f(\zeta )|$, then $|c_n|\le M_{\rho }{\rho }^{-n}$. These bounds justify differentiating the power series any fixed number of times on a smaller closed disc, by comparison with convergent differentiated geometric series. Evaluation at $a$ gives $f^{(n)}(a)=n!c_n$. Since every $|z-a|<R$ belongs to some such smaller disc, the formula holds throughout the stated domain.

The identity theorem says that two holomorphic functions on a connected open subset of $\mathbb{C}$ are equal everywhere if their equality set has an accumulation point in that subset. Equivalently, a holomorphic function with an interior accumulation point of zeros is identically zero there.

The proposed Taylor coefficients at $i$, including the specified value for $n=0$, giveThe function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ and has exactly the given derivatives, so it supplies existence. Any other admissible holomorphic function has the same Taylor series on $|z-i|<1$. The punctured plane is connected, so the identity theorem gives uniqueness:

Write $M=|f(z_0)|$. The local maximum property gives a disk around $z_0$ on which $|f|\le M$, contained in the unit disk. If $M=0$, then $f=0$ on that neighbourhood, and the identity theorem makes $f$ zero on the connected unit disk.

Otherwise multiply $f$ by the unit complex number $\overline{f(z_0)}/M$, obtaining a holomorphic function $g$ with $g(z_0)=M$ and $|g|\le M$ nearby. Applying the Cauchy integral formula on any sufficiently small circle of radius $r$ gives the mean value propertyTake real parts. The continuous function $M-\mathrm{Re}g$ is nonnegative on the circle and has integral zero, so it vanishes everywhere there. Since $|g|\le M$, having real part $M$ forces imaginary part zero. Thus $g=M$ throughout the circle. The zeros of $g-M$ have an accumulation point inside its domain, so the identity theorem gives $f$ constant on the entire unit disk. This proves the local maximum modulus principle directly from the Cauchy integral formula.

Suppose, towards a contradiction, that for some $r,w,\epsilon$ the punctured-disk image misses the open disk of radius $\epsilon$ about $w$. Thenis holomorphic and bounded by $1/\epsilon$ on $0<|z|<r$. Part (i) extends it to a holomorphic function $H$ on the full disk. If $H(0)\ne 0$, then $w+1/H$ extends $f$ holomorphically at zero, making zero a removable singularity.

If $H(0)=0$, the function $H$ is not identically zero, since it is nonzero on the punctured disk. The Taylor series and identity theorem give a finite zero order $k\ge 1$, so $H(z)=z^{k}q(z)$ with $q$ holomorphic and $q(0)\ne 0$. Hencehas a pole of order $k$. Both alternatives contradict an essential singularity. ConsequentlyThis is the Casorati-Weierstrass theorem: every complex value can be approximated arbitrarily closely in every punctured neighbourhood of an essential singularity.