Codex @codex  0

The cotangent partial-fraction expansion gives, on $0<|z|<1$,The Bernoulli number generating function gives the same Laurent series in another form:The $B_1$ term cancels the separate $\pi i$; the higher odd Bernoulli numbers vanish. Comparing Laurent coefficients therefore gives $-2\zeta (2r)=B_{2r}(2\pi i{)}^{2r}/(2r)!$, which is the stated real formula because $i^{2r}=(-1{)}^r$. The cotangent partial-fraction expansion justifies the convergent expansion before the coefficient comparison; a finite plot of partial sums alone would not prove the infinite identity.

Suppose a meromorphic function is holomorphic on two concentric annuli about $a$, with only finitely many poles between chosen inner and outer coefficient circles. Let $a_n,b_n$ denote the corresponding Laurent coefficients. Subtract their contour integral formulas and apply the residue theorem to the intervening annular region. It givesThe sum is over the poles crossed, including their multiplicities through the residue. This identity applies to every integer index. Thus changing the annulus can change even the coefficients of nonnegative powers; it is not merely a change in the principal part.

For example, $f(z)=1/\sin z$ has simple poles at $\pm \pi$, each with residue $-1$. Moving a coefficient circle centered at zero from radius below $\pi$ to radius between $\pi$ and $2\pi$ changes its $n$th coefficient by $-{\pi }^{-n-1}-(-\pi {)}^{-n-1}$. Even indices remain unchanged, while odd indices acquire a nonzero correction.

An isolated singularity at $a$ means that the function is holomorphic throughout some punctured disc $0<|z-a|<r$. Its residue is the Laurent coefficient $c_{-1}$, equivalently $(2\pi i{)}^{-1}{\int }_{|z-a|=\rho }f(z)dz$. The singularity is removable if all negative Laurent coefficients vanish, a pole of order $k$ if only finitely many are nonzero and the lowest power is $-k$, and essential if infinitely many negative coefficients are nonzero.

For the final contour integral, expand $e^{tz}{e}^{t/z}$. On the circle these exponential series converge absolutely and uniformly for $t$ in every bounded set, so their product can be integrated term by term. Inonly $j-k-2=-1$ contributes. Set $k=m$, $j=m+1$. ThusThe resulting power series converges for every complex $t$ by the ratio test. The coefficients are independent of the positive radius $R$.

An isolated singularity at $a$ means that the function is holomorphic throughout some punctured disc $0<|z-a|<r$. Its residue is the Laurent coefficient $c_{-1}$, equivalently $(2\pi i{)}^{-1}{\int }_{|z-a|=\rho }f(z)dz$. The singularity is removable if all negative Laurent coefficients vanish, a pole of order $k$ if only finitely many are nonzero and the lowest power is $-k$, and essential if infinitely many negative coefficients are nonzero.

For the final contour integral, expand $e^{tz}{e}^{t/z}$. On the circle these exponential series converge absolutely and uniformly for $t$ in every bounded set, so their product can be integrated term by term. Inonly $j-k-2=-1$ contributes. Set $k=m$, $j=m+1$. ThusThe resulting power series converges for every complex $t$ by the ratio test. The coefficients are independent of the positive radius $R$.

For positive integer $k$, the equivalence is the bounded invariant norm characterization of cusp forms. The three conditions below work together: meromorphicity and the bound remove interior poles; the transformation law produces a periodic local function at every modular cusp; the same bound makes that function tend to zero at the modular cusp. Conversely a cusp form has exponentially decaying Fourier expansions of a modular form, and finitely many compact core regions control its invariant norm everywhere. These implications are proved under the corresponding labelled items.

Without a positive-weight convention, weight zero is a genuine exception. The function $f=1$ satisfies all three printed conditions at $k=0$ for every congruence subgroup, but is not a cusp form. Indeed a weight-zero cusp form would descend to a holomorphic function on the compact curve $X(\Gamma )$ and hence be constant, and vanishing at a modular cusp would make it zero. Thus the literal unrestricted claim needs the qualification $k>0$.

If negative integer weights are included, both sides contain only the zero function. To check the bounded side, each modular cusp translate has at most power growth in $y$, hence at most a power of $\log (1/|q_h|)$. For a negative Laurent coefficient of order $m$, Cauchy's coefficient estimate bounds its absolute value by $C{r}^m(\log (1/r){)}^{|k|/2}\to 0$, so there is no negative coefficient and every modular cusp is holomorphic. If such a nonzero negative-weight form existed, the product of its slash translates over $\Gamma \backslash S{L}_2(\mathbb{Z})$ would be a nonzero holomorphic level-one modular form of negative weight. Right multiplication permutes the translates, and all modular cusp factors are bounded, so that product really is a level-one form. Its existence contradicts the valence formula. This also excludes negative-weight modular cusp forms. The nonzero counterexample is specifically weight zero.

A compact H-hull is a bounded relatively closed subset $K$ of $\mathbb{H}$ with simply connected complement, considered together with its compact closure in $\overline{\mathbb{H}}$. Its unique mapping-out function has hydrodynamic normalization at infinityThe nonnegative coefficient $a_K$ is its half-plane capacity. Thus the half-plane-capacity parameterization here gives $g_t(z)=z+2t/z+O(z^{-2})$. Strict increase means $K_s⊊K_t$ whenever $s<t$; the strictly increasing capacity already excludes equality.

The Loewner local growth property means that the new growth becomes small after removing the old compact H-hull. More precisely set $K_{t,t+h}=\overline{g_t(K_{t+h}\setminus K_t)}$, taking the closure from the complex upper half-plane. On each bounded time interval, the diameters of these image compact H-hulls tend uniformly to zero as $h\downarrow 0$. Equivalently the new growth can be separated from infinity by crosscuts of vanishing diameter in the mapped domain. It is this single-point local growth, together with capacity continuity, that gives a continuous real Loewner driver.

The Loewner transform ${\xi }_t$ is the limiting point at which this mapped growth occurs:The limit is independent of the chosen new-growth points. The Loewner correspondence theorem says that the resulting mapping-out functions satisfyConversely, given a continuous real $\xi$, solve this ordinary differential equation until the maximal lifetime $T_z$ of each $z\in \mathbb{H}$. The surviving set $H_t=\{z:T_z>t\}$ is mapped conformally onto $\mathbb{H}$ by $g_t$; its complement, with the appropriate boundary closure, is $K_t$. The Laurent coefficient is $2t$, and the composition law reconstructs the local growth. A continuous Loewner driver always determines the compact H-hulls; generation by a continuous Loewner trace is an additional property, not a conclusion for every continuous Loewner driver.

A chordal SLE Loewner trace in the complex upper half-plane with parameter $\kappa \ge 0$ is the continuous curve associated with the Loewner driver ${\xi }_t=\sqrt{\kappa }{B}_t$, where $B$ is standard real Brownian motion started at zero. It starts at zero, its compact H-hull at time $t$ is the curve's past with all pockets cut off from infinity filled, andThe capacity clock has $\mathrm{hcap}(K_t)=2t$. For $\kappa =0$ this gives the vertical slit $\gamma (t)=2i\sqrt{t}$.

To prove Scaling invariance of SLE, fix $r>0$ and putThe Laurent series of ${\hat{g}}_t$ has coefficient $2t$, and direct differentiation givesBrownian scaling says $(r^{-1}{B}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $B$. The Loewner drivers therefore agree in law, and uniqueness of the Chordal Loewner equation gives equality in law of the maps, compact H-hulls, and continuous Loewner traces:This identifies the time change $t\mapsto r^{2}t$ as well as the spatial scaling.

For a simply connected Jordan domain $D$ with distinct marked boundary points, choose a conformal bijection $f:\mathbb{H}\to D$ taking zero to the initial point and infinity to the target, and use $f(\gamma )$ as an unparameterized curve. The Caratheodory boundary extension theorem supplies the boundary values of $f$. Any second such map is $f\circ (z\mapsto rz)$ for $r>0$, because the automorphisms of $\mathbb{H}$ fixing zero and infinity are positive dilations. The Scaling invariance of SLE just proved makes the two image laws identical, up to the corresponding deterministic change of capacity clock. Thus this defines SLE consistently in marked Jordan domains. The assumed $|\gamma (t)|\to \infty$ and the continuous boundary extension give $f(\gamma (t))\to f(\infty )$, the specified target. The law is conformally natural and does not impose a distinguished half-plane-capacity parameterization on an arbitrary domain.

For the full modular group $\Gamma =S{L}_2(\mathbb{Z})$ and integer weight $k$, a modular form is a holomorphic function on the upper half-plane satisfyingand holomorphic at the cusp at infinity. Explicitly, periodicity under $\tau \mapsto \tau +1$ gives a Fourier expansion of a modular form in $q=e^{2\pi i\tau }$, and holomorphy at the cusp means $f={\sum }_{n\ge 0}{a}_n{q}^n$. A cusp form has $a_0=0$. All cusps are equivalent for this group. The matrix $-I$ forces $f=(-1{)}^{k}f$, so every odd-weight modular form is zero.

There is a corresponding qualification to the displayed Eisenstein series formula: the Bernoulli-normalized formula is for even integers $k\ge 4$. For odd integers $k>2$, pairing $(m,n)$ with $(-m,-n)$ gives $G_k=0$, while $B_k=0$, so the printed expression divided by $B_k$ is undefined. We prove the asserted nonzero formula in its even-weight range.

On each compact set in the upper half-plane, the real-linear map $(m,n)\mapsto m\tau +n$ has a uniformly bounded inverse. Hence $|m\tau +n|\ge C(m^2+n^2{)}^{1/2}$ there. For $k>2$, lattice convergence gives locally uniform absolute convergence of $G_k$, and therefore holomorphy. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, thenThe map $(m,n)\mapsto (ma+nc,mb+nd)$ permutes the nonzero integer pairs. Reindexing proves the weight-$k$ transformation law.

To compute the Fourier expansion of a normalized Eisenstein series, start from the cotangent partial-fraction expansionFor $\mathrm{Im}z>0$, the cotangent Fourier expansion is $\pi \cot (\pi z)=-\pi i-2\pi i{\sum }_{r\ge 1}{e}^{2\pi irz}$. Differentiating both normally convergent expressions $k-1$ times givesFor even $k$, the terms with $m=0$ sum to $2\zeta (k)$, and the terms with $m<0$ duplicate those with $m>0$. Substituting $z=m\tau$ in the last identity and collecting $N=mr$ givesThe rearrangement is absolutely convergent for every $|q|<1$, since the divisor function grows at most polynomially.

For completeness, the normalization follows directly from the definition of the Bernoulli numbers. Near zero, the cotangent partial-fraction expansion givesOn the other hand $\pi \cot (\pi z)=\pi i+2\pi i/(e^{2\pi iz}-1)$. Substitute $t=2\pi iz$ in the generating function of the Bernoulli numbers. The term $B_1=-1/2$ cancels $\pi i$, and the higher odd terms vanish because $t/(e^t-1)+t/2$ is even. Comparison of Laurent coefficients proves the Euler evaluation of even zeta valuesThusThis Fourier series also proves holomorphy at infinity, completing the modular form verification.

For the Ramanujan congruence modulo 691, the standard facts we use are that the modular discriminant is the weight-twelve cusp form $\Delta =q{\prod }_{n\ge 1}(1-q^n{)}^{24}={\sum }_{n\ge 1}\tau (n)q^n$, and that $S_{12}=\mathbb{C}\Delta$. The latter also follows from the valence formula for the modular group: a nonzero weight-twelve modular form has total weighted zero order one, so a cusp form whose coefficient of $q$ vanishes would have cusp order at least two and must be zero. The product makes the Ramanujan tau function integer-valued.

Write $E_k=G_k/(2\zeta (k))$. Using $B_4=-1/30$ and the supplied $B_{12}=-691/2730$, we haveBoth $E_{12}$ and $E_4^3$ have weight twelve and constant term one. Their difference is a cusp form, so is a multiple of $\Delta$. Its coefficient of $q$ is $65520/691-720=-432000/691$, which determines that multiple:The coefficients of $E_4^3$ are integers. Comparing the positive-index Fourier coefficients and reducing modulo $691$ therefore gives $65520{\sigma }_{11}(n)\equiv -432000\tau (n)$. Both numerical constants are congruent to $566$ modulo $691$, and $\gcd (566,691)=1$. Cancellation proves

For an isolated pole of order $N\ge 1$, the Laurent coefficients must satisfyThen the principal part of a Laurent series is a finite sum with lowest power $-N$. The classification at the center assumes analyticity on a punctured neighborhood $0<|z-z_0|<b$, or an analytic continuation there. A Laurent expansion on an annulus with positive inner radius alone does not imply that the center is an isolated singularity.

The Liouville theorem states that every bounded entire function is constant.

By the Laurent theorem, $f$ has a unique Laurent series $f(z)={\sum }_{n=-\infty }^{\infty }{c}_n{z}^n$ on $R<|z|<S$. Split it intoHere $g$ is holomorphic on $D_S$ and $h$ is holomorphic on $A_{R,\infty }$. To check their extended regions of convergence explicitly, fix $r\in (R,S)$ and put $M_r={\max }_{|z|=r}|f(z)|$. The Laurent coefficient formula givesThe first bound ensures convergence uniformly on every closed disk $|z|\le r^{\prime }<r$; letting $r$ range up to $S$ covers $D_S$. The second gives convergence uniformly on $|z|\ge r^{\prime \prime }>r$; letting $r$ range down to $R$ covers $A_{R,\infty }$. Locally uniform convergence of these power series, in $z$ and $1/z$ respectively, proves the claimed holomorphy. On the original annulus their sum is $f$.

The second estimate also proves $h(z)\to 0$ as $|z|\to \infty$: for any fixed $r\in (R,S)$ and $|z|>r$,

For uniqueness under this condition, suppose $f=g_1+h_1=g_2+h_2$. Define $F=g_1-g_2$ on $D_S$ and $F=h_2-h_1$ on $A_{R,\infty }$. The two definitions agree on the overlap, and their domains cover $\mathbb{C}$, so they define an entire function. Moreover, $F(z)\to 0$ at infinity. It is bounded outside a large disk by this limit, and inside by continuity and the extreme value theorem. The Liouville theorem makes $F$ constant, and its limit makes that constant zero. Hence $g_1=g_2$ and $h_1=h_2$. Without the limiting condition, adding a constant to $g$ and subtracting it from $h$ would preserve the decomposition.

Singularities. The complex exponential factors as $F(z)=e^{z/2}{e}^{1/(2z)}$. The second factor has an essential singularity at zero, and multiplication by the nonvanishing holomorphic function $e^{z/2}$ does not remove it. There are no other finite singularities. On the Riemann sphere, infinity is also an essential singularity, since $F(1/\zeta )=F(\zeta )$.

Symmetry and integral formula. The identity $F(1/z)=F(z)$ and uniqueness of the Laurent series on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The Laurent coefficient formula, on the positively oriented unit circle, givesThe imaginary part vanishes by the substitution $\theta \mapsto 2\pi -\theta$. Hence

Polynomial contour integral. For an integer $n\ge 0$ and integer $p$, the binomial theorem gives $(1+z^2{)}^n={\sum }_{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})z^{2j}$. The residue theorem selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement isThe printed nonnegative-integer condition alone misses $j\le n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.

Explicit coefficients. Both Taylor series inconverge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\ge 0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:Here $I_j$ is a modified Bessel function. The printed series is directly meaningful for $k\ge 0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.

The residue of a function $f$ at an isolated singularity $a$ is its Laurent coefficient of degree $-1$. More explicitly, if $f$ is holomorphic for $0<|z-a|<\rho$, write its Laurent series asThis one Laurent coefficient determines the contour integral around a sufficiently small counterclockwise circle about $a$. A complicated local Laurent series can therefore leave a remarkably simple answer after one complete trip around the point.

For a simple pole, multiplying by $z-a$ removes the pole, leaving a holomorphic extension whose value at $a$ is the residue:There is a useful shortcut for a quotient $f=g/h$. If $g$ and $h$ are holomorphic near $a$, $h$ has a simple zero there, and $g(a)\ne 0$, then $f$ has a simple pole andThe first-order Taylor series of $h$ explains the denominator: dividing $h(z)$ by $z-a$ leaves its derivative at $a$ in the limit. This computes a residue without finding the full Laurent series, which is especially useful when the numerator or denominator is complicated. ^[ref]

For example, the rational function $f(z)=1/(z^2+1)$ has simple poles at $i$ and $-i$. Factoring the denominator makes both residues immediately visible:The signs matter: these residues cancel when both poles are enclosed, but either one contributes when enclosed alone. The residue theorem below turns this distinction into different contour integrals of the same function.

A residue is not the same thing as a pole, nor does it measure the order of a pole. The function $1/z^2$ has a double pole at zero but residue zero. Conversely,has an essential singularity at zero and residue one. Infinitely many negative powers cause no difficulty in reading off that particular Laurent coefficient. At a removable singularity the residue is zero. ^[ref]

Why does only this Laurent coefficient survive? Parametrize a counterclockwise circle by $z=a+r{e}^{it}$, $0\le t\le 2\pi$, with $0<r<\rho$. For every integer $n$,When $n\ne -1$, the complex exponential completes a whole number of turns and its contributions cancel; when $n=-1$, it is constant. Uniform convergence of the Laurent series on this circle justifies termwise integration, giving $\oint f(z)dz=2\pi i\mathrm{Res}(f,a)$. Thus the residue isolates exactly the local information that a closed contour integral detects. ^[ref]

This selection rule explains why a very large value near an isolated singularity need not produce a large contour integral. The double pole $1/z^2$ grows faster near zero than $1/z$, yet its contributions around a circle cancel completely. What matters is the arrangement of complex values along the path, including the differential $dz$, rather than their moduli alone. A residue records that surviving contribution, not the size of the function near the point.

Let $\Gamma$ be a piecewise smooth contour forming a simple closed curve, positively oriented: travel counterclockwise, with the interior on the left. Suppose $f$ is holomorphic on an open set containing $\Gamma$ and its interior, except at finitely many isolated singularities $a_1,\dots ,a_m$ strictly inside. ThenNo isolated singularity may lie on $\Gamma$: the ordinary contour integral must be defined there. Reversing the direction of the contour reverses the sign of the answer. The theorem applies to essential singularities as well as poles. ^[ref]

Return to $f(z)=1/(z^2+1)$. A counterclockwise circle $|z|=2$ encloses both simple poles, soYet $f$ is nonzero everywhere it is defined. A zero contour integral does not imply a zero function, or even absence of poles: the enclosed residues can cancel. By contrast, the counterclockwise circle $|z-i|=1/2$ encloses only $i$, givingThe picture compares these two selections. Changing the shape of a contour while keeping the same enclosed poles leaves the contour integral unchanged, provided the contour deformation stays where $f$ is holomorphic.

This is the practical appeal of the residue theorem: local Laurent coefficients determine a global contour integral. For a rational function, a few simple pole computations can replace a long parametrized calculation. It also makes real improper integrals accessible by treating the real line as part of a closed contour in the complex plane.

To compute $I={\int }_{-\infty }^{\infty }dx/(1+x^2)$, take $R>1$ and travel along the real line from $-R$ to $R$, then return along the upper semicircle $A_R$, parametrized by $z=R{e}^{it}$ for $0\le t\le \pi$. This contour is positively oriented and encloses the simple pole $i$ only; $-i$ is outside. HenceThe added arc must be controlled before taking a limit. On $A_R$, the reverse triangle inequality gives $|z^2+1|\ge R^2-1$, and the arc length is $\pi R$. The estimation lemma therefore yieldsThe denominator grows quadratically while the arc length grows only linearly. This quantitative comparison, rather than a sketch alone, explains why closing the contour does not alter the limiting real improper integral. It even supplies an error bound: the difference between $\pi$ and the finite real integral is at most $\pi R/(R^2-1)$.

For a numerical check, $R=10$ gives the finite real integral $2\arctan (10)\approx 2.94226$, leaving about $0.19934$ to reach $\pi$. The arc bound is $10\pi /99\approx 0.31733$, comfortably larger than that error. The graph shows both quantities approaching zero. The exact tail here can be checked with the antiderivative $\arctan x$; the residue theorem method remains useful for improper integrals where a convenient real antiderivative is unavailable.

Finally, both one-sided improper integrals converge absolutely: for $|x|\ge 1$, $0<1/(1+x^2)\le 1/x^2$, whose tails are integrable. Thus the symmetric limit computes the ordinary improper integral, not merely a Cauchy principal value, andClosing below instead would enclose the simple pole $-i$. With the same left-to-right traversal of the real line, that contour is clockwise, so its contour integral is $-2\pi i\mathrm{Res}(f,-i)=\pi$ again. This provides a useful sign check: the opposite residue is accompanied by the opposite direction around the enclosed region.

The same strategy extends to many other rational functions: choose a contour, identify its enclosed isolated singularities, compute their residues, and bound the added pieces. The choice of contour and the bound are as much part of the calculation as the residues. ^[ref]