lacunary - Mathnotes

Residue Integration

Suppose f(z) has a singularity at z=z0 inside a simple closed curve C but is otherwise analytic on C and inside C.

Then, f(z) has a @laurent-series

f(z)=n=0an(zz0)n+b1zz0+b2(zz0)2+

that converges for all points near z=z0 (except at z=z0 itself,) in some domain of the form 0<|zz0|<R.

The coefficient b1 of the first negative power 1/(zz0) of this @laurent-series is given by the @Cauchy-integral-formula as

b1=12piiCf(z)dz.

Now, we can use this to find the value of the integral without using any of the integral formulas:

Cf(z)dz=2πib1.

This is a CCW integral around a simple closed path C that contains z=z0 in its interior (but no other singularities of f(z) on or inside C.)

Definition: Residue \@{residue}

Given a convergent @laurent-series

f(z)=n=0an(zz0)n+b1zz0+b2(zz0)2+,

the coefficient b1 of the first negative power of 1/(zz0) is called the residue of f(z) at z=z0. It is denoted by

b1=Resz=z0f(z).

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Residue Formulas

Instead of finding the @laurent-series, we can use these handy formulas.

Simple pole at z0

Theorem: First Simple Pole Residue Formula \@{first-simple-pole-residue-formula}

Resz=z0f(z)=b1=limzz0(zz0)f(z).

Theorem: Second Simple Pole Residue Formula \@{second-simple-pole-residue-formula}

Resz=z0f(z)=Resz=z0p(z)q(z)=p(z0)q(z0).

Poles of any Order at z0

Theorem \@{residue-at-mth-order-pole}

Resz=z0f(z)=1(m1)!limzz0{dm1dzm1[(zz0)mf(z)]}.

Remark \@{remark-5}

For second order poles (m=2), this gives

Resz=z0f(z)=limzz0{[(zz0)2f(z)]}.

Several Singularities Inside the Contour

Theorem: Residue Theorem \@{residue-theorem}

Let f(z) be analytic inside a simple closed path C and on C, except for finitely many @singular-points z1,z2,,zk inside C. Then,

Cf(z)dz=2πij=1kResz=zjf(z).

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