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A simple zero of an analytic function has a local factorization $f(z)=(z-z_0)u(z)$, where $u$ is an analytic function and $u(z_0)=f^{\prime }(z_0)\ne 0$. Since $1/u$ is also an analytic function near $z_0$, its Taylor series givesThe nonzero coefficient of $(z-z_0{)}^{-1}$ proves that this is a simple pole, withMultiplying by the Taylor series $g(z)=g(z_0)+O(z-z_0)$ similarly givesThe assumption $g(z_0)\ne 0$ ensures that the pole is not canceled. In either formula the residue is precisely the coefficient of $(z-z_0{)}^{-1}$ in the Laurent series.

Clear the apparent negative power of $z$ to obtain the rational functionLet $\Delta =\sqrt{a^2-1}>0$. Its denominator factors as $(z-z_{-})(z-z_{+})$, withThese are distinct simple poles. Since $(a-\Delta )(a+\Delta )=1$ and $a+\Delta >1$, we have $0<a-\Delta <1$: only $z_{-}$ lies inside the unit circle. The apparent singularity at $z=0$ in the original expression is a removable singularity, with the continued value $h(0)=0$; it is not another pole.

Using the residue formula from part (a), with denominator derivative $2z-2ia$, givesThere are no other finite singularities of the continued rational function.

First use $\sin \theta /(a-\sin \theta )=a/(a-\sin \theta )-1$. If $J={\int }_{-\pi }^{\pi }(a-\sin \theta {)}^{-1}d\theta$, the required integral is $aJ-2\pi$.

Set $z=e^{i\theta }$. The unit circle is traversed counterclockwise once, $d\theta =dz/(iz)$, and $\sin \theta =(z-z^{-1})/(2i)$. ThusThe function $h(z)/z=-2i/(z^2-2iaz-1)$ is regular at zero. Its only enclosed pole is $z_{-}$ from part (b), and its residue isBy the residue theorem, $J=(1/i)2\pi i/\Delta =2\pi /\Delta$. ThereforeAs a sign and scale check, this is positive and behaves as $\pi /a^2$ for large $a$, in agreement with expanding the integrand and integrating its first nonzero, ${\sin }^2\theta /a^2$, term.

Expand the exponential in inverse powers of $z$:This is a Laurent series, not a Taylor series, and converges throughout $0<|z|<\infty$. Every displayed coefficient of a negative power is nonzero, so its principal part has infinitely many terms. ThusThe residue is the coefficient of $z^{-1}$:

For a positively oriented simple closed contour, the residue theorem states that the integral of a function holomorphic on and inside the contour except at finitely many isolated interior singularities is $2\pi i$ times the sum of its residues. There must be no singularity on the contour.

Factor the denominator as $(z^2+z+1)(z^2-z+1)$. For $R>1$, the upper semicircle encloses the two simple polesThe other two poles are in the lower half-plane. Since $D^{\prime }(z)=4z^3+2z=2z(2z^2+1)$,Writing $h=\sqrt{3}/2$ and using $1/a=1/2-ih$, $1/b=-1/2-ih$, their numerator difference simplifies toThus for every $R>1$ the closed-contour integral isTo remove the arc, on $C_R$ we have $|e^{iz}|=e^{-\mathrm{Im}z}\le 1$ and $|z^4+z^2+1|\ge R^4-R^2-1$. For sufficiently large $R$, the complex line integral estimate givesThe real-line integral is absolutely convergent. Its imaginary part is zero by oddness of $\sin x/(x^4+x^2+1)$. Therefore

At $\Omega =\kappa$, the inverse-transform denominator is $-(\omega -i\kappa {)}^2$. For $t>0$, its upper-half-plane double-pole residue is the derivative of the exponential:It is also the continuous limit of both answers above as $\beta \to 0$ or $\eta \to 0$. This critically damped response tends to zero with a linear factor multiplying the exponential.

In all three cases, for $t<0$ close the contour in the lower half-plane, where $e^{i\omega t}$ decays. There are no poles inside, soAt $t=0$ its continuous value is zero and $G^{\prime }(0^{+})=1$, producing the unit derivative jump required by $(D_t^2+2\kappa D_t+{\Omega }^2)G=\delta (t)$. Thus all three formulas are causal Green's functions and all decay to zero at late times.

In the Laurent seriesabout a pole of order $m$, the residue is the coefficientwhere the circle is traversed anticlockwise. Equivalently, the residue is $\frac{1}{(m-1)!}{\frac{d^{m-1}}{d{z}^{m-1}}[(z-z_0{)}^{m}f(z)]|}_{z=z_0}$.

The Taylor series of the entire function $g$ givesThe coefficient of $(z-z_0{)}^{-1}$, and therefore the residue, isIf $g(z_0)=0$, the apparent pole is a removable singularity and the residue is zero.

For the simple-contour form of the residue theorem, let $C$ be a positively oriented, simple, closed, piecewise smooth contour. Assume $f$ is holomorphic on a neighbourhood of $C$ and its interior except at finitely many isolated singularities $z_1,\dots ,z_s$, all strictly inside $C$, with none on $C$. ThenThe more general version weights each residue by the contour's winding number, with the usual requirement that the contour be homologous to zero in the domain of analyticity containing it and the nonsingular region.

For the requested application, assume $f$ is holomorphic on a neighbourhood of the contour and its closed interior, and $z_0$ is strictly inside it. Its Taylor series showsOnly the second term contributes a residue. The residue theorem therefore proves the derivative form of the Cauchy integral formula:

Use $z=e^{i\theta }$, so $\sin \theta =(z-z^{-1})/(2i)$ and $d\theta =dz/(iz)$. The unit circle is traversed anticlockwise, and the integral becomesOnly the simple pole $z=i/3$ lies inside the contour. Its residue isHence the residue theorem gives

Close the real segment $[-R,R]$ by an anticlockwise semicircle in the upper half-plane. For $F(z)=1/[(z^2+1)(z^2+4)]$, the contour integral over the arc is $O(R^{-3})$: its length is $\pi R$, whereas $|z^2+1|\ge R^2-1$ and $|z^2+4|\ge R^2-4$. The enclosed simple poles are $i$ and $2i$, with residuesLetting $R\to \infty$ in the residue theorem yields

Factor the denominator as $(z-z_0)(z+z_0)$. Since $z_0\ne 0$, its zeros are distinct, and the potential simple poles and their residues areFor example, the first follows by taking ${\lim }_{z\to z_0}(z-z_0)f(z)$. If the numerator vanishes at either denominator zero, the singularity there is removable and the corresponding residue is zero. Thus the data identify the possible poles, not an unconditional pole at a point canceled by $g$.

If $f$ is analytic on and inside a positively oriented simple closed contour except for finitely many isolated poles $z_j$ strictly inside it, then the residue theorem statesThere must be no singularity on the contour. More generally each residue is multiplied by the contour's winding number about its point.

Let $n\ge 2$ be even and integrate $f(z)=1/(1+z^n)$ around an upper-half-plane semicircle of radius $R>1$ with the real segment traversed from $-R$ to $R$. There are no real poles, and the $n/2$ upper-half-plane poles areEach is simple. Since $z_j^n=-1$,On the semicircle the integral magnitude is at most $\pi R/(R^n-1)$, which tends to zero. The real integral is absolutely convergent because its tail is of order $|x{|}^{-n}$. ThusEvaluate the finite geometric series, writing $\theta =\pi /n$:ConsequentlyFor $n=2$ this reduces to $\pi$, as a check on the contour orientation and residue signs.

Put $s=1/n$, so $0<s<1$. For the complex exponentiation $z^s$, choose $-\pi /2<\arg z<3\pi /2$: the branch cut is the negative imaginary axis. Integrate $z^s/(a^2+z^2)$ around the upper half-plane semicircle, with a small semicircular indentation excluding the origin. The only enclosed simple pole is $z=ia$, with residueThe large arc is $O(R^{s-1})$ and the small arc is $O({ϵ}^{s+1})$, so both vanish. On the negative real segment, $z^s=e^{i\pi s}|z{|}^s$; accounting for the direction of traversal, the two real segments therefore sum to $(1+e^{i\pi s})I$, where $I$ is the desired positive-axis integral. The residue theorem givesSince $1+e^{i\pi s}=2e^{i\pi s/2}\cos (\pi s/2)$,The endpoint estimates also prove convergence. More generally the same calculation is the fractional-power integral from an upper-half-plane contour, valid for $-1<s<1$.

Set $z=e^{i\theta }$. Then $d\theta =dz/(iz)$ and $\tan \theta =(z^2-1)/[i(z^2+1)]$, so the requested integral isThe denominator factors asFor $0<a<1$, its two inner roots are $z=\pm r$, $r=\sqrt{(1-a)/(1+a)}<1$, and the other two roots have modulus $1/r>1$. There is also a simple pole at zero. Their residues areThe latter values follow by differentiating the quadratic product, or from the supplied hint. The sum of the enclosed residues isThe residue theorem therefore givesAt the angles where the tangent is undefined, the original reciprocal has a removable zero limit. Its continuous extension gives the same integral; the transformed rational integrand has no pole on the unit circle.

Rotations commute with one another, while reflection conjugation sends $R^j$ to $R^{-j}$. Thus the rotational conjugacy classes are pairs $\{R^j,R^{-j}\}$, except when the two entries coincide. For reflections, direct use of the relations givesIf $n$ is odd, multiplication by two permutes residues modulo $n$, so all reflections are conjugate. The complete class list isTheir sizes are $1$, then $(n-1)/2$ classes of size two, then one class of size $n$, giving $(n+3)/2$ classes and $2n$ elements in total.

If $n$ is even, the half-turn $R^{n/2}$ is central and forms its own singleton. Conjugation changes a reflection exponent by an even amount or negates it, so it preserves parity and reaches every exponent of that parity. The list isThere are $n/2+3$ classes; their sizes are $1,1$, then $n/2-1$ pairs, then two classes of size $n/2$. Geometrically the two reflection classes correspond to vertex-axis and edge-axis reflections, in either order depending on the choice of $m$. These are the conjugacy classes of a finite dihedral group.

For $z\ne 0$, singularities occur when $1/z=n\pi$, with $n\ne 0$. Each is a simple pole, since the substitution has nonzero derivative there:More explicitly, $1/z-n\pi \sim -(z-z_n)/z_n^2$, so $\cot (1/z)\sim -z_n^2/(z-z_n)$ and its residue is $-1/(n^2{\pi }^2)$.

These poles accumulate at zero. Consequently $\boxed{z=0\text{ is a non-isolated singularity}}$. The removable/pole/essential classification for isolated singularities cannot be applied at that accumulation point; it is not an isolated essential singularity.

The zeros of $\sin z$ are $z=n\pi$ for $n\in \mathbb{Z}$, all simple, and $\cos (n\pi )=(-1{)}^n\ne 0$. Hence the finite singularities of $\cot z$ are exactlyLocally $\cot z\sim 1/(z-n\pi )$, so each pole has residue $1$.

For this real integral choose $0<\arg z<2\pi$, so the upper bank of the positive real axis has $z^{1/4}=x^{1/4}>0$. Use a counterclockwise keyhole contour with inner radius $\epsilon$ and outer radius $R$, enclosing $\pm i$. The upper-bank contribution tends to $J$. On the lower bank, $z^{-1/4}=e^{-i\pi /2}{x}^{-1/4}=-i{x}^{-1/4}$, and the reversed orientation makes its contribution $+iJ$.

The large arc is $O(R^{-5/4})$ because $g=O(R^{-9/4})$, while the small arc is $O({\epsilon }^{3/4})$. Both vanish, givingThe residues on this branch areThe residue theorem yieldsThereforeThe branch cut in (a) was arbitrary; here placing it on the integration axis makes the bank contributions directly proportional to the desired positive-real integral.