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An isolated singularity at $a$ means that $f$ is holomorphic in some punctured disk $0<|z-a|<r$, while $a$ itself is an excluded point or a point where the given function is not holomorphic. There are no other singularities arbitrarily close to $a$. By Laurent's theorem, the function has a unique Laurent series on that punctured disk:The negative-power terms form its principal part of a Laurent series.

An isolated singularity at $a$ means that $f$ is holomorphic in some punctured disk $0<|z-a|<r$, while $a$ itself is an excluded point or a point where the given function is not holomorphic. There are no other singularities arbitrarily close to $a$. By Laurent's theorem, the function has a unique Laurent series on that punctured disk:The negative-power terms form its principal part of a Laurent series.

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.

Use the local coordinate $w=1/z$ at infinity. For positive integer $N$,The negative powers of this local coordinate are precisely $m=0,\dots ,N-1$. The singular part at infinity is thereforeInfinity is a pole at infinity of order $N$. In the original $z$ variable these singular terms are positive powers; it is the negative powers of $w=1/z$ that define the local principal part of a Laurent series. For $N=0$ the point at infinity is regular, and for negative integer $N$ it is a zero rather than a pole.

Laurent's theorem states that a function holomorphic on $R<|z-a|<S$ has a unique expansionconverging uniformly on every compact subannulus. Its coefficients arewith counterclockwise orientation. The value is independent of the chosen intermediate radius. The nonnegative powers form a power series convergent inside the outer radius, and the negative powers form the principal part of a Laurent series.

At a simple pole, write $f(z)=c/(z-a)+h(z)$ with $h$ holomorphic near $a$ and $c$ its residue. On the arc $z=a+ϵe^{it}$, the singular term integrates exactly to $ic(\beta -\alpha )$. The remainder has absolute integral at most $ϵ(\beta -\alpha )\sup |h|$, which tends to zero. Reversing orientation reverses the sign. In particular a clockwise upper semicircle contributes $-i\pi c$ in the shrinking limit.

A higher-order pole need not have such a limit: $(z-a{)}^{-2}$ has residue zero but integral $2/ϵ$ on a positively oriented upper semicircle. Thus the simple-pole assumption controls all the potentially divergent terms of the principal part of a Laurent series.