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Multiply the exponential power series by the two prefactors:Collecting powers yields the Laurent seriesIts first terms are $z+1+3/(2z)+7/(6z^2)+13/(24z^3)+\cdots$. Infinitely many negative-power coefficients are nonzero, so $\boxed{z=0\text{ is an essential singularity}}$. The residue is the coefficient of $z^{-1}$:The phrase "power series" here refers to a Laurent series; no ordinary Taylor series about an essential singularity exists.

The Fourier transform of a derivative gives $(-{\omega }^2-1)\tilde{y}=-\tilde{f}$. For $\lambda =1$,This selects the solution decaying at both time infinities. Indeed, the inverse transform of $1/(1+{\omega }^2)$ is the one-dimensional modified Helmholtz Green function $K(t)=e^{-|t|}/2$, and the convolution theorem gives $y=K\ast f$. The printed equation has homogeneous solutions $e^t,e^{-t}$, so these two decay conditions matter; no causal initial-value condition was specified.

For $t<0$, evaluate Fourier inversion by closing the contour in the lower half-plane, where $e^{i\omega t}$ decays. The contour is clockwise. The enclosed simple poles and residues of $\tilde{y}$ areThus inversion equals $-i$ times the sum of the residues of $e^{i\omega t}\tilde{y}$, givingAs a check, applying $d^2/d{t}^2-1$ gives $-e^t\cos 2t$. The convolution also gives $y(t)=e^{-t}[\frac{1}{8}+\frac{1}{5}\cos t+\frac{2}{5}\sin t]$ for $t\ge 0$; the two pieces have matching values $13/40$ and matching first derivatives $3/40$ at zero.

A simple pole at $z_0$ means that $f$ is holomorphic in a punctured neighbourhood and $(z-z_0)f(z)$ extends holomorphically to a nonzero value at $z_0$. Equivalently its Laurent series begins with $a_{-1}/(z-z_0)$ and has no higher negative powers. The residue isIn particular, if $f=g/h$ with $g(z_0)\ne 0$, $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, this becomes $\mathrm{Res}(f,z_0)=g(z_0)/h^{\prime }(z_0)$. The associated residue theorem isfor a positively oriented simple contour with no poles on it.

The exponential numerator is entire and never zero. The denominator vanishes precisely when $e^{2z}=-1$, givingAt these points $\sinh z_n=i(-1{)}^n\ne 0$, so every zero of the denominator is simple and hence every singularity is a simple pole. Using the residue formula in (b),

Use the counterclockwise rectangular contour with vertices $-R,R,R+i\pi ,-R+i\pi$. It encloses just the simple pole $i\pi /2$, whose residue is $e^{k\pi /2}/i$. Let $I_R={\int }_{-R}^R{e}^{-ikx}/\cosh xdx$. The identitymakes the top edge, traversed right to left, contribute $e^{k\pi }{I}_R$. On either vertical edge,Thus the sum of the vertical integrals tends to zero, with absolute value at most $2\pi e^{|k|\pi }/\sinh R$. The residue theorem and $R\to \infty$ now giveThe argument works for every fixed real $k$, including zero and negative values, and gives the Fourier transform of the hyperbolic secant with the convention $\tilde{f}(k)=\int e^{-ikx}f(x)dx$.

For fixed $z\notin \mathbb{Z}$, integratecounterclockwise around the specified square of half-side $R_N=N+1/2$, choosing $N>|z|$. The integer poles inside it are $-N,\dots ,N$, and their residues are $1/(n^2-z^2)$ because $\pi \cot (\pi w)$ has residue one at each integer.

The two other possible poles are $w=z$ and $w=-z$. Their residues are each $\pi \cot (\pi z)/(2z)$, hence sum to $\pi \cot (\pi z)/z$. This formula still applies when the numerator vanishes and a possible pole is removable. The residue theorem givesOn the square, $|w|\ge R_N$, its perimeter is $8R_N$, and the supplied cotangent bound givesThe series converges absolutely by part (b), so its symmetric partial sums have the same limit as the full sum. ThereforeAt half-integers the cotangent form makes the value zero explicit; the reciprocal-tangent notation is understood by its continuous extension there. The cotangent contour summation rests on the contour estimate, not only on formally matching poles.

Write the Laurent series about $a$ as ${\sum }_{k\in \mathbb{Z}}{c}_k(z-a{)}^k$. The isolated singularity is:

These cases exhaust the possible principal parts. The residue isFor the first example, $\sin z=z+O(z^3)$ givesThus zero is a double pole, with residue zero. For the other example,so the ratio is $-z/2+O(z^2)$. Its singularity is removable, with extended value zero and residue zero. The numerator's zero has higher order than the denominator's, so there is no pole.

A simple zero means $h(a)=0$ and $h^{\prime }(a)\ne 0$. The local factorization is $h(z)=(z-a)h_1(z)$, where $h_1$ is holomorphic and $h_1(a)=h^{\prime }(a)\ne 0$. ThusThe numerator is holomorphic near $a$, so its constant Taylor coefficient is the coefficient of $(z-a{)}^{-1}$ in the quotient. ThereforeIf $g(a)=0$, the possible pole cancels and the residue is zero; the formula does not require $g/h$ actually to have a pole. This is the residue formula for a quotient at a simple denominator zero.

Consider the counterclockwise unit circle and setThe parametrization $z=e^{i\theta }$ givesThe imaginary part is zero: replacing $\theta$ by $2\pi -\theta$ conjugates the integrand, since $n$ is an integer and $x$ is real. The real part is precisely the defining Bessel integral, so $I=J_n(x)$.

Both exponential series converge absolutely and uniformly on the circle. Their product may therefore be integrated term by term:Uniform absolute convergence can also be checked by summing the absolute coefficient bounds, which gives $e^{|x|}$. The residue is obtained exactly when $n-1+j-k=-1$, or $k=n+j$. The Cauchy coefficient formula consequently yieldsThis proves the Bessel function of the first kind series directly from its integral, including the sign and factorial normalization.

One standard form of Cauchy's residue theorem is the following. Let $\Gamma$ be a positively oriented contour that is a simple closed curve, and let $f$ be a holomorphic function on an open set containing $\Gamma$ and its interior except for finitely many interior isolated singularities $z_1,\dots ,z_m$. There are no singularities on $\Gamma$. ThenThe residue $\mathrm{Res}(f,z_j)$ is the coefficient of $(z-z_j{)}^{-1}$ in the Laurent series at $z_j$. Each interior isolated singularity contributes once because the winding number of this positively oriented contour about each such point is one. Reversing the orientation of a contour reverses the sign. This form of Cauchy's residue theorem applies to the rectangle below.

Put $D(z)={\sinh }^{2}z+{\cosh }^{2}a$ and $S=\sinh (2a)>0$. Traverse the rectangle $-R\to R\to R+\pi i\to -R+\pi i\to -R$ with positive orientation of a contour, taking $R>a$. Part (i) shows that the only enclosed poles of $f(z)=z/D(z)$ areThey are simple poles becauseThe residue formula for a quotient at a simple denominator zero givesTheir imaginary terms cancel and their residues sum to $-2a/S$. Thus Cauchy's residue theorem gives

The choice of numerator $z$ makes the horizontal sides useful. The hyperbolic sine satisfies $\sinh (z+\pi i)=-\sinh z$, so $D(x+\pi i)=D(x)$. Consequently the lower and upper contour integrals, with their actual orientations, add toIt remains to justify discarding both vertical sides. On either side $z=\pm R+iy$, $0\le y\le \pi$, the reverse triangle inequality and part (b)(ii) implyThis lower bound is positive for sufficiently large $R$. Also $|z|\le R+\pi$ and each vertical side has length $\pi$. The estimation lemma bounds the absolute value of the sum $V_R$ of the vertical contributions byThe real improper integral converges absolutely, since its integrand is continuous on the real axis and decays like $4e^{-2|x|}$ at infinity. Combining all four sides and taking $R\to \infty$ givesThereforeAn independent real-variable check uses the hyperbolic tangent substitution $u=\tanh x$: the integral becomes ${\int }_{-1}^{1}du/({\cosh }^{2}a-u^2{\sinh }^{2}a)$. Its antiderivative uses the inverse hyperbolic tangent, giving $2\mathrm{arctanh}(\tanh a)/(\sinh a\cosh a)=4a/\sinh (2a)$. As $a\downarrow 0$, this tends to $2$, agreeing with ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{2}xdx=2$ and checking the factor of two.

The residue $\mathrm{res}(f;a)$ is the coefficient $c_{-1}$ of $(z-a{)}^{-1}$ in the Laurent series at the isolated singularity. Equivalently it is $(2\pi i{)}^{-1}$ times the integral around a sufficiently small counterclockwise circle enclosing $a$.

The residue theorem states that if $f$ is holomorphic on a neighborhood of a positively oriented simple closed contour and its interior except for finitely many isolated singularities $a_j$ inside, thenNo singularity may lie on the integration contour. A more general contour weights each residue by its winding number.

The Taylor expansions giveand hence $f(z)=b/z+iab+O(z)$. Since $b>0$, zero is a simple pole with residue $b$. It is not a removable singularity, despite the vanishing numerator, because the denominator has a zero of order two.

For $z\ne 0$, the denominator vanishes exactly when $1/z=(2k+1)\pi i$. Thus setEach denominator zero is simple, sinceEvery $z_k$ is purely imaginary and nonzero. The identity $\sin (iy)=i\sinh y$ for real $y$ shows $\sin z_k\ne 0$, so there is no cancellation. Each $z_k$ is consequently a simple pole, with residue $z_k^2\sin z_k$.

These poles accumulate at zero. ThereforeIn particular, zero has no punctured neighbourhood on which the function is holomorphic, so it cannot be classified as a removable singularity, a pole, or an isolated essential singularity. There are no other finite singularities.

For $a\ne b$, integratealong the contour consisting of $[-R,R]$ and the upper semicircle, with positive orientation of a contour and $R>\max \{a,b\}$. Its enclosed simple poles are $ia$ and $ib$, with residuesOn the upper semicircle, $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$. The reverse triangle inequality gives $|z^2+a^2|\ge R^2-a^2$ and $|z^2+b^2|\ge R^2-b^2$, so the estimation lemma givesThe integral along the real axis is absolutely convergent: the denominator stays nonzero on bounded real intervals, and the integrand is $O(x^{-4})$ at infinity. The residue theorem therefore givesTake real parts. The real integrand is an even function, so

To include $a=b$, fix $a>0$ and let $b\to a$. For $a/2\le b\le 3a/2$, the absolute value of the integrand is bounded byan integrable function on $[0,\infty )$. The dominated convergence theorem justifies passing the limit under the integral. If $h(t)=e^{-t}/t$, then $h^{\prime }(a)=-e^{-a}(a+1)/a^2$, andHence

Use the principal logarithm $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z<\pi$, whose branch cut is the negative real axis. For this choice,has three simple poles away from the cut: $z=1$, $z=ia$, and $z=-ia$. The pole at $1$, caused by $\mathrm{Log}1=0$, must be included.

Choose $0<\epsilon <\min \{1,a\}$ and $R>\max \{1,a\}$, and use a positively oriented keyhole contour around the negative real axis: the large circle is traversed counterclockwise, the upper bank from $-R$ to $-\epsilon$, the small circle clockwise, and the lower bank from $-\epsilon$ to $-R$. Formally, use radial segments at arguments $\pm (\pi -\eta )$, then let $\eta \downarrow 0$ with $R,\epsilon$ fixed. The function has continuous limiting values on each compact bank, which justifies this limit.

Put $L=\log a$. The residues areThe first follows from $\mathrm{Log}z=(z-1)+O((z-1{)}^2)$; the other two follow from the simple zeros of $z^2+a^2$. Their sum isThese expressions are valid also for $a=1$, since $L\pm i\pi /2$ never vanish for real $L$.

On the outer circle, $|\mathrm{Log}z|\ge \log R$ for $R>1$ and $|z^2+a^2|\ge R^2-a^2$. On the inner circle, $|\mathrm{Log}z|\ge |\log \epsilon |$ and $|z^2+a^2|\ge a^2-{\epsilon }^2$. Thus the estimation lemma bounds the two arc integrals byrespectively, as $R\to \infty$ and $\epsilon \downarrow 0$.

On the upper bank, $z=-x+i0$ gives $\mathrm{Log}z=\log x+i\pi$; on the lower bank it gives $\log x-i\pi$. Accounting for the opposite orientations, their combined integral isThe real improper integral converges: its integrand is bounded near zero by $1/(a^2{\pi }^2)$ and at infinity by $1/({\pi }^2{x}^2)$. Taking the arc limits in the residue theorem gives $-2\pi iI=2\pi i\Sigma$, henceThe final subtraction is exactly the contribution of the pole at $1$.

Let $F=e^g$. It is holomorphic and nowhere zero on a punctured neighbourhood of $a$, and the chain rule givesIf $F$ had a pole of order $n\ge 1$, part (c) would imply ${\mathrm{Res}}_a(g^{\prime })=-n$. But the derivative of a holomorphic function on a punctured disc has residue zero. Indeed, write its Laurent seriesTermwise differentiation is valid on every smaller closed annulus. The only possible $(z-a{)}^{-1}$ term comes from $k=0$ and has coefficient zero. Thusa contradiction. Therefore $\boxed{e^g\text{ cannot have a pole at }a}$. This proves an exponential has no isolated pole, including when $g$ has an essential singularity; no choice of a holomorphic logarithm on the entire punctured disc was assumed.

The simple zero assumption gives the local factorizationwhere $k$ is holomorphic. In a small disc $k$ stays nonzero, so $q=g/k$ is holomorphic there. Expanding $q$ to first order,The residue is thereforeIf $g(a)=0$, the quotient has a removable singularity rather than a simple pole, and the same formula correctly gives zero. This is the residue formula for a quotient at a simple denominator zero.

The degree $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower theorem states that for intermediate fields $F\subseteq E\subseteq K$ in a finite extension,If the degree is prime, one factor must be one, forcing $E=K$ or $E=F$; no strictly intermediate field exists.

Let $K$ have 32 elements. Its prime subfield is ${\mathbb{F}}_2$ and $[K:{\mathbb{F}}_2]=5$. Every element outside that subfield has minimal polynomial of degree five, by the tower theorem. Also every member of $K$ satisfies $x^{32}=x$: nonzero elements form a group of order 31. Hence $x^{32}-x$ splits over $K$, with distinct roots because its derivative is $-1=1$ in characteristic two.

Every irreducible degree-five polynomial over ${\mathbb{F}}_2$ divides $x^{32}-x$. Indeed its quotient field has 32 elements, and its residue class of $x$ satisfies that equation. Thus every such polynomial splits into five distinct roots in $K$. Conversely each of the 30 elements outside ${\mathbb{F}}_2$ has one such minimal polynomial, and different minimal polynomials have disjoint root sets. Consequently the number isThere is no extra leading-coefficient factor, since every nonzero leading coefficient over ${\mathbb{F}}_2$ is one.

Euler's theorem states that $a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$ whenever $\gcd (a,m)=1$, where $\varphi (m)$ counts the invertible residue classes. For each prime-power divisor, the same theorem gives $a^{\varphi (p_i^{e_i})}\equiv 1(\mathrm{mod}{p}_i^{e_i})$. Since the exponent defined in the source is a multiple of every one of these exponents, raising that congruence to the corresponding integer power givesThe prime powers are pairwise coprime, so their product divides $a^{\lambda (m)}-1$, proving the congruence modulo $m$.

For the displayed composite integer,Thus every integer coprime to 561 satisfies $\boxed{a^{560}\equiv 1(\mathrm{mod}561)}$. This is a Carmichael number. The source's prime-power least common multiple need not be the minimal universal exponent for all moduli; for powers of two its definition can exceed the usual Carmichael function. No minimality claim is needed here.