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The input is even, so its Fourier transform is real and even. Decomposing the cosine into complex exponentials and integrating over its support givesCombining the fractions and using the angle addition formula yieldsThe apparent poles at $k=\pm 1$ are removable singularities; either the integral or the sine-quotient limit gives $\tilde{f}(\pm 1)=a+\sin (2a)/2$.

If $\cos a\ne 0$, expansion of the denominator givesThe oscillatory envelope is therefore $O(|k{|}^{-1})$. At $a=\pi /2+n\pi$, $n=0,1,2,\dots$, the endpoint values vanish and the first term cancels:This gives an $O(|k{|}^{-2})$ envelope. At general $a$ the truncation creates jumps in the function; at these special values it leaves the function continuous but creates jumps in its first derivative. Each additional degree of endpoint smoothness gives another inverse power, as explained by Fourier decay from a derivative jump. The exact example is the Fourier transform of a truncated cosine; isolated zeros of its oscillatory factors do not change the envelope order.

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.

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.

The only possible finite singularities are $z_{\pm }=\pm i\pi /2$, the zeros of $4z^2+{\pi }^2$. They are simple zeros of the denominator, because its derivative $8z_{\pm }$ is nonzero. They are also simple zeros of the numerator: $\cosh z_{\pm }=0$ and $\sinh z_{\pm }=\pm i\ne 0$.

The common simple factors cancel, so both points are removable singularities, with extension valuesThere are no other finite singularities; after inserting these two values the function is entire.

The expression is $F(z)=e^{z^2+1/z}=e^{z^2}{e}^{1/z}$, which is holomorphic for every $z\ne 0$. At zero,has infinitely many nonzero negative-power terms, so it has an essential singularity. Multiplication by the nonvanishing entire function $e^{z^2}$ preserves that classification: if $F$ had a removable singularity or a pole, multiplying by $e^{-z^2}$ would give the same type for $e^{1/z}$, a contradiction. Hence

An isolated singularity at $a$ is a removable singularity when there is a holomorphic function $F$ on a full disc $B(a,r)$ agreeing with $f$ away from $a$. Equivalently,The Riemann removable singularity theorem also characterizes this by boundedness of $f$ on a sufficiently small punctured disc. A preassigned incorrect value at $a$ is replaced by $L$.

A function has an isolated singularity at $a$ if, for some $r>0$, it is defined and holomorphic on the punctured disc $0<|z-a|<r$, while a holomorphic extension through $a$ has not been assumed. In particular, its domain omits $a$, or its assigned value there fails to give such an extension. The key local condition isThe convention includes removable singularities, even when a suitable extension would eliminate the missing point.

A function has a pole of order $n$ at $a$ when, on a punctured neighbourhood,Equivalently, $(z-a{)}^{n}f(z)$ has a removable singularity whose extension is nonzero at $a$. After shrinking the neighbourhood so $h\ne 0$, its reciprocal $1/f$ extends with a zero of order $n$. In particular, $|f(z)|\to \infty$ at a pole.

We prove the Casorati-Weierstrass theorem by contradiction. Fix an allowed $\rho$. If the stated image were not a dense set in $\mathbb{C}$, there would be a disc $B(w,ϵ)$ disjoint from it. HenceThe holomorphic function $h(z)=1/(f(z)-w)$ is bounded by $1/ϵ$ on that punctured disc. By the permitted Riemann removable singularity theorem, it extends to a holomorphic function $H$ on the full disc.

If $H(a)\ne 0$, then $w+1/H$ is holomorphic near $a$ and agrees with $f$ away from $a$. This makes $a$ a removable singularity of $f$.

If $H(a)=0$, then $H$ is not identically zero, since $h$ is nowhere zero away from $a$. Its Taylor series therefore has a finite first nonzero coefficient: otherwise the identity theorem would make it identically zero. Write $H(z)=(z-a{)}^{m}K(z)$ with $m\ge 1$ and $K(a)\ne 0$. Thenhas a pole of order $m$ at $a$, since multiplying by $(z-a{)}^m$ gives the nonzero limit $1/K(a)$.

Both cases contradict an essential singularity. Thus

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 product rule applied at the interpolation node $x_k$ givesThus Lagrange interpolation gives, away from the interpolation nodes,Apply the same Lagrange interpolation formula to the constant polynomial one. Since its interpolant is itself,Dividing the preceding two identities proves the barycentric interpolation formulaThe denominator is exactly $1/w(x)$, so is nonzero away from the interpolation nodes. At $x=x_j$, multiply numerator and denominator by $x-x_j$ and take the limit: their values tend to $w_j{f}_j$ and $w_j$, respectively. The removable singularity therefore has value $f_j$; in numerical evaluation return the stored value directly at an interpolation node.

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 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.

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.

A holomorphic function bounded on a punctured neighbourhood extends holomorphically across the puncture.

This criterion characterizes a removable singularity of a holomorphic function; the criterion is a theorem about the singularity type.

For $a>0$, let $g(x)=f(ax)$. Whenever the Mellin transform of $f$ is absolutely convergent,The Mellin fundamental strip is unchanged: the corresponding integrals of absolute values differ by the finite positive factor $a^{-\sigma }$. The change of variables $y=ax$ gives $x^{s-1}dx=a^{-s}{y}^{s-1}dy$, proving the rule. For $s=\sigma +i\tau$, the transform's absolute value changes by $a^{-\sigma }$, while the factor $e^{-i\tau \log a}$ changes its phase. On the line $\sigma =0$, wherever absolute convergence holds, the transform's absolute value is preserved.

For the Mellin transform of an interval indicator, replacing $1_{[1,4]}(x)$ by $1_{[1,4]}(2x)=1_{[1/2,2]}(x)$ givesAt the removable singularity $s=0$, both sides have value $\log 4$. Scaling leaves the width measured by the natural logarithm unchanged. Likewise, the exponential function $e^{-ax}$ has Mellin transform $a^{-s}\Gamma (s)$ on $\mathrm{Re}s>0$.

A square-root branch point has ramification index $2$: one circuit switches local branches and two restore the original branch. Both zero and infinity are square-root branch points of $z^{-1/2}$, since the coordinate $u=1/z$ changes the latter local expression to $u^{1/2}$. A finite limiting value at a branch point does not imply a removable singularity.