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Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

The apparent singularity at zero cancels: the Taylor series of the sine givesThus zero is a removable singularity, with extension $f(0)=1/2$. At $z=-2$, the other terms are holomorphic, whereas $(z+2{)}^{-1}$ has a simple pole. ThereforeThere are no other finite singularities. If “all” includes the point at infinity on the Riemann sphere, it is an essential singularity: $f(1/w)=w/(1+2w)+w^2\sin (1/w)-w$ has infinitely many negative powers in its Laurent series at $w=0$.

The geometric series for $(2+z{)}^{-1}$ converges for $|z|<2$, while the cancelled sine terms have an everywhere convergent Taylor series. Hence the requested Laurent series has no negative powers:For example,The extension through zero also makes this a Taylor series. The pole at $-2$ fixes its radius of convergence at two.

Let $f$ be holomorphic on a neighbourhood of a positively oriented, piecewise smooth simple closed curve and its interior except at finitely many isolated singularities $a_j$ strictly inside. The residue theorem statesThe residue is the coefficient of $(z-a_j{)}^{-1}$ in the local Laurent series. No singularity may lie on the contour. For more general closed contours, each residue is multiplied by the corresponding winding number.

Use the principal complex square root, $\sqrt{z}=\sqrt{r}{e}^{i\theta /2}$ with $-\pi <\theta <\pi$. For $F(z)=e^{z-\sqrt{z}}/z$, the local expansion on this holomorphic branch isThe fractional power prevents a single-valued Laurent series in a punctured disc. In particular, zero is a branch point, not an isolated singularity classified as a pole. Merely observing that $zF(z)\to 1$ on the slit plane would not establish a simple pole: $zF(z)$ does not extend as a holomorphic function through a whole disc around zero.

The negative real half-line is a chosen branch cut. Its upper and lower boundary values differ because $\sqrt{-x+i0}=i\sqrt{x}$ and $\sqrt{-x-i0}=-i\sqrt{x}$ for $x>0$. Continuation once around zero changes the sign of the complex square root, and hence changes $F$ in general. Other branch cuts are possible; this one defines the principal holomorphic branch.

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.

Singularities and cuts. There are four simple poles $z_j=e^{i(\pi /4+j\pi /2)}$, $j=0,1,2,3$, because the roots of $1+z^4$ are simple and do not coincide with $\pm 1$. At $z=\pm 1$, the local behavior is a nonzero constant times $(z\mp 1{)}^{-1/2}$: these are square-root branch points. Infinity is not a branch point; on either branch $f(z)=O(z^{-5})$, so it extends there with a zero of order five. Three valid choices of branch cuts are the segment $[-1,1]$, the two rays $(-\infty ,-1]\cup [1,\infty )$ joined through infinity, or a semicircular arc joining $-1$ to $1$ and avoiding the poles. A circuit about either endpoint changes the sign, while a circuit about both endpoints does not.

Branch and residues. On $\mathbb{C}\setminus [-1,1]$, let $s(z)=\sqrt{z^2-1}$ be the branch with $s(z)/z\to 1$ at infinity, and define $\sqrt{1-z^2}=-is(z)$. Its upper boundary value on $(-1,1)$ is positive. The residue at a simple pole $z_j$ isFor fully explicit values, putIn counterclockwise order $z_0,z_1,z_2,z_3$, the values of $s$ are $A+iB,-A+iB,-A-iB,A-iB$. Since $A^2+B^2=\sqrt{2}$, substitution yieldsA branch point has no ordinary residue, since it has no single-valued Laurent series on a punctured neighborhood. At infinity the residue is zero.

Contour evaluation. The original contour goes counterclockwise around a large circle and clockwise round the slit, with its upper lip directed to the right. The large-circle contribution is $O(R^{-4})$. Small endpoint arcs contribute $O({ϵ}^{1/2})$. The artificial slit extension from $1$ to $R$ has equal boundary values on both sides, so its contributions cancel. On $(-1,1)$ the two boundary values of $f$ are opposite. Thus, with $I$ denoting the requested real integral, the residue theorem givesTherefore

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

Expanding the exponential function produces the Laurent seriesIn particular its residue is $a_{-1}=n^{n+1}/(n+1)!$.

For a positively oriented circle of any radius $R>0$, the residue theorem givesOnly the $z^{-1}$ coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

Use the branch cut $[-1,1]$ and the branch of a multivalued function normalized by $\sqrt{z^2-1}\sim z$. The Laurent series from the preceding solution has $z^{-1}$ coefficient $3/8$. Consequently every sufficiently large positively oriented circle givesAlthough $f$ grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

The Laurent series isIts principal part contains infinitely many nonzero negative powers, so zero is an essential singularity. The first three displayed terms are not an asymptotic truncation as $z\to 0$; they are the beginning of the convergent Laurent series in powers of $1/z$.

The coefficient of $z^{-1}$ in the Laurent series is $-1$. The residue theorem therefore gives, for the standard positive orientation,The residue theorem applies to an isolated essential singularity as well as to a pole.

A function holomorphic outside a disk has a pole of positive integer order $N$ at infinity when its reciprocal-coordinate representative $f(1/w)$ has a pole of that order at $w=0$. Equivalently $f(z)=z^{N}g(1/z)$ with $g$ holomorphic and nonzero at zero. The Laurent series about infinity is interpreted in this coordinate: finitely many positive powers of $z$ form its principal part. For example, $z^N{e}^{1/z}=z^N+z^{N-1}+\cdots$ has a pole of order $N$ at infinity despite its essential singularity at zero.

The principal part consists of all negative-power terms of a Laurent series about its center. On a punctured neighborhood of an isolated singularity it is zero for a removable singularity, a nonzero finite sum for a pole, and an infinite series for an essential singularity. A pole's order is the largest $N$ with $a_{-N}\ne 0$. At infinity use the local variable $w=1/z$: negative powers of $w$ correspond to positive powers of $z$. A Laurent series on an annulus with strictly positive inner radius may have a principal part without the function being defined close enough to the center to classify an isolated singularity there.

At a point $x$ of a Riemann surface, the principal part of a meromorphic function is the negative-degree part of its Laurent series in a local coordinate. It is equivalently a germ in the quotient ${\mathcal{K}}_x/{\mathcal{O}}_x$.

The principal part at a pole is the finite sum of negative-degree terms of its Laurent series. Prescribing such parts is the problem solved by the Mittag-Leffler theorem.

For holomorphic $N,D$ with $N(z_0)D(z_0)\ne 0$, write the quotient as $h(z)/(z-z_0)$ where $h=N/D$. Its Taylor series immediately gives the displayed Laurent series. The residue is $h(z_0)$ and the constant regular term is $h^{\prime }(z_0)$; the latter requires differentiating both numerator and remaining denominator. Higher regular coefficients follow from the higher Taylor coefficients of $h$.

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.