Residue (complex analysis)

Contour integration

Contour integral of a monomial

Computing the residue of a monomial

${\displaystyle {\oint }_C{z}^{k}dz}$

makes most residue computations easy to do. Since path integral computations are homotopy invariant, we will let ${\displaystyle C}$ be the circle with radius ${\displaystyle 1}$ going counter clockwise. Then, using the change of coordinates ${\displaystyle z\to e^{i\theta }}$ we find that

${\displaystyle dz\to d(e^{i\theta })=i{e}^{i\theta }d\theta }$

hence this integral now reads as

${\displaystyle {\oint }_C{z}^{k}dz={\int }_0^{2\pi }i{e}^{i(k+1)\theta }d\theta =\{\begin{array}{ll}2\pi i& \text{if }k=-1,\\ 0& \text{otherwise}.\end{array}}$

Thus, the residue of ⁠${\displaystyle z^k}$⁠ is ⁠${\displaystyle 1}$⁠ if integer ⁠${\displaystyle k=-1}$⁠ and ⁠${\displaystyle 0}$⁠ otherwise.

Generalization to Laurent series

If a function is expressed as a Laurent series expansion around ⁠${\displaystyle c}$⁠ as follows: $${\displaystyle f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-c{)}^n.}$$ Then, the residue at the point ⁠${\displaystyle c}$⁠ is calculated as: $${\displaystyle \mathrm{Res}(f,c)=\frac{1}{2\pi i}{\oint }_{\gamma }f(z)dz=\frac{1}{2\pi i}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\oint }_{\gamma }{a}_n(z-c{)}^{n}dz=a_{-1}}$$ using the results from contour integral of a monomial for counter clockwise contour integral ⁠${\displaystyle \gamma }$⁠ around a point ⁠${\displaystyle c}$⁠. Hence, if a Laurent series representation of a function exists around ⁠${\displaystyle c}$⁠, then its residue around ⁠${\displaystyle c}$⁠ is known by the coefficient of the term ⁠${\displaystyle (z-c{)}^{-1}}$⁠.

Application in residue theorem

For a meromorphic function ⁠${\displaystyle f}$⁠, with a finite set of singularities within a positively oriented simple closed curve ${\displaystyle C}$ which does not pass through any singularity, the value of the contour integral is given according to residue theorem, as: $${\displaystyle {\oint }_{C}f(z)dz=2\pi i\sum _{k=1}^n\mathrm{I}(C,a_k)\mathrm{Res}(f,a_k).}$$ where ⁠${\displaystyle \mathrm{I}(C,a_k)}$⁠, the winding number, is ⁠${\displaystyle 1}$⁠ if ⁠${\displaystyle a_k}$⁠ is in the interior of ⁠${\displaystyle C}$⁠ and ⁠${\displaystyle 0}$⁠ if not, simplifying to: $${\displaystyle {\oint }_{\gamma }f(z)dz=2\pi i\sum \mathrm{Res}(f,a_k)}$$ where ⁠${\displaystyle a_k}$⁠ are all isolated singularities within the contour ⁠${\displaystyle C}$⁠.

Calculation of residues

Suppose a punctured disk ⁠${\displaystyle D=\{z:0<|z-c|<R\}}$⁠ in the complex plane is given and ⁠${\displaystyle f}$⁠ is a holomorphic function defined (at least) on ⁠${\displaystyle D}$⁠. The residue ⁠${\displaystyle \mathrm{Res}(f,c)}$⁠ of ⁠${\displaystyle f}$⁠ at ⁠${\displaystyle c}$⁠ is the coefficient ⁠${\displaystyle a_{-1}}$⁠ of ⁠${\displaystyle (z-c{)}^{-1}}$⁠ in the Laurent series expansion of ⁠${\displaystyle f}$⁠ around ⁠${\displaystyle c}$⁠. Various methods exist for calculating this value, and the choice of which method to use depends on the function in question, and on the nature of the singularity.

According to the residue theorem, we have:

${\displaystyle \mathrm{Res}(f,c)=\frac{1}{2\pi i}{\oint }_{\gamma }f(z)dz}$

where ⁠${\displaystyle \gamma }$⁠ traces out a circle around ⁠${\displaystyle c}$⁠ in a counterclockwise manner and does not pass through or contain other singularities within it. We may choose the path ⁠${\displaystyle \gamma }$⁠ to be a circle of radius ⁠${\displaystyle \epsilon }$⁠ around ⁠${\displaystyle c}$⁠. Since ⁠${\displaystyle \epsilon }$⁠ can be as small as we desire it can be made to contain only the singularity of ⁠${\displaystyle c}$⁠ due to nature of isolated singularities. This may be used for calculation in cases where the integral can be calculated directly, but it is usually the case that residues are used to simplify calculation of integrals, and not the other way around.

Removable singularities

If the function ⁠${\displaystyle f}$⁠ can be continued to a holomorphic function on the whole disk ⁠${\displaystyle |y-c|<R}$⁠, then ⁠${\displaystyle \mathrm{Res}(f,c)=0}$⁠. The converse is not in general true.

Simple poles

If ⁠${\displaystyle c}$⁠ is a simple pole of ⁠${\displaystyle f}$⁠, the residue of ⁠${\displaystyle f}$⁠ is given by:

${\displaystyle \mathrm{Res}(f,c)=\underset{z\to c}{lim}(z-c)f(z).}$

If that limit does not exist, then ⁠${\displaystyle f}$⁠ instead has an essential singularity at ⁠${\displaystyle c}$⁠. If the limit is ⁠${\displaystyle 0}$⁠, then ⁠${\displaystyle f}$⁠ is either analytic at ⁠${\displaystyle c}$⁠ or has a removable singularity there. If the limit is equal to infinity, then the order of the pole is higher than ⁠${\displaystyle 1}$⁠.

It may be that the function ⁠${\displaystyle f}$⁠ can be expressed as a quotient of two functions, ⁠${\displaystyle f(z)=g(z)/h(z)}$⁠, where ⁠${\displaystyle g}$⁠ and ⁠${\displaystyle h}$⁠ are holomorphic functions in a neighbourhood of ⁠${\displaystyle c}$⁠, with ⁠${\displaystyle h(c)=0}$⁠ and ⁠${\displaystyle h^{\prime }(c)\ne 0}$⁠. In such a case, L'Hôpital's rule can be used to simplify the above formula to:

${\displaystyle \begin{array}{cc}\mathrm{Res}(f,c)& =\underset{z\to c}{lim}(z-c)f(z)=\underset{z\to c}{lim}\frac{zg(z)-cg(z)}{h(z)}\\ & =\underset{z\to c}{lim}\frac{g(z)+z{g}^{\prime }(z)-c{g}^{\prime }(z)}{h^{\prime }(z)}=\frac{g(c)}{h^{\prime }(c)}.\end{array}}$

Limit formula for higher-order poles

More generally, if ⁠${\displaystyle c}$⁠ is a pole of order ⁠${\displaystyle p}$⁠, then the residue of ⁠${\displaystyle f}$⁠ around ⁠${\displaystyle z=c}$⁠ can be found by the formula:

${\displaystyle \mathrm{Res}(f,c)=\frac{1}{(p-1)!}\underset{z\to c}{lim}\frac{d^{p-1}}{d{z}^{p-1}}\left((z-c{)}^{p}f(z)\right).}$

This formula can be very useful in determining the residues for low-order poles. For higher-order poles, the calculations can become unmanageable, and series expansion is usually easier. For essential singularities, no such simple formula exists, and residues must usually be taken directly from series expansions.

Residue at infinity

In general, the residue at infinity is defined as:

${\displaystyle \mathrm{Res}(f(z),\mathrm{\infty })=-\mathrm{Res}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right).}$

If the following condition is met:

${\displaystyle \underset{|z|\to \mathrm{\infty }}{lim}f(z)=0,}$

then the residue at infinity can be computed using the following formula:

${\displaystyle \mathrm{Res}(f,\mathrm{\infty })=-\underset{|z|\to \mathrm{\infty }}{lim}zf(z).}$

If instead

${\displaystyle \underset{|z|\to \mathrm{\infty }}{lim}f(z)=c\ne 0,}$

then the residue at infinity is

${\displaystyle \mathrm{Res}(f,\mathrm{\infty })=\underset{|z|\to \mathrm{\infty }}{lim}{z}^2{f}^{\prime }(z).}$

For functions that are meromorphic on the entire complex plane with finitely many singularities, the sum of the residues at the (necessarily) isolated singularities plus the residue at infinity is zero, which gives:

${\displaystyle \mathrm{Res}(f(z),\mathrm{\infty })=-\sum _k\mathrm{Res}(f(z),a_k).}$

Series methods

If parts or all of a function can be expanded into a Taylor series or Laurent series, which may be possible if the parts or the whole of the function has a standard series expansion, then calculating the residue is significantly simpler than by other methods. The residue of the function is simply given by the coefficient of ⁠${\displaystyle (z-c{)}^{-1}}$⁠ in the Laurent series expansion of the function.

Examples

Residue from series expansion

Example 1

As an example, consider the contour integral

${\displaystyle {\oint }_C\frac{e^z}{z^5}dz}$

where ⁠${\displaystyle C}$⁠ is some simple closed curve about ⁠${\displaystyle 0}$⁠.

Let us evaluate this integral using a standard convergence result about integration by series. Substituting the Taylor series for ⁠${\displaystyle e^z}$⁠ into the integrand, the integral becomes

${\displaystyle {\oint }_C\frac{1}{z^5}\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}+\frac{z^6}{6!}+\cdots \right)dz.}$

Let us bring the term in ⁠${\displaystyle 1/z^5}$⁠ into the series. The contour integral of the series then writes

${\displaystyle \begin{array}{cc}& {\oint }_C\left(\frac{1}{z^5}+\frac{z}{z^5}+\frac{z^2}{2!z^5}+\frac{z^3}{3!z^5}+\frac{z^4}{4!z^5}+\frac{z^5}{5!z^5}+\frac{z^6}{6!z^5}+\cdots \right)dz\\ =& {\oint }_C\left(\frac{1}{z^5}+\frac{1}{z^4}+\frac{1}{2!z^3}+\frac{1}{3!z^2}+\frac{1}{4!z}+\frac{1}{5!}+\frac{z}{6!}+\cdots \right)dz.\end{array}}$

Since the series converges uniformly on the support of the integration path, we are allowed to exchange integration and summation. The series of the path integrals then collapses to a much simpler form because of the previous computation. So now the integral around ⁠${\displaystyle C}$⁠ of every other term not in the form ⁠${\displaystyle c{z}^{-1}}$⁠ is zero, and the integral is reduced to

${\displaystyle {\oint }_C\frac{1}{4!z}dz=\frac{1}{4!}{\oint }_C\frac{1}{z}dz=\frac{1}{4!}(2\pi i)=\frac{\pi i}{12}.}$

The value 1/4! is the residue of ⁠${\displaystyle e^z/z^5}$⁠ at ⁠${\displaystyle z=0}$⁠, and is denoted

${\displaystyle {\mathrm{Res}}_0\frac{e^z}{z^5},\text{ or }{\mathrm{Res}}_{z=0}\frac{e^z}{z^5},\text{ or }\mathrm{Res}(f,0)\text{ for }f=\frac{e^z}{z^5}.}$

Example 2

As a second example, consider calculating the residues at the singularities of the function$${\displaystyle f(z)=\frac{\sin z}{z^2-z}}$$which may be used to calculate certain contour integrals. This function appears to have a singularity at ⁠${\displaystyle z=0}$⁠, but if one factorizes the denominator and thus writes the function as $${\displaystyle f(z)=\frac{\sin z}{z(z-1)}}$$ it is apparent that the singularity at ⁠${\displaystyle z=0}$⁠ is a removable singularity and then the residue at ⁠${\displaystyle z=0}$⁠ is therefore ⁠${\displaystyle 0}$⁠. The only other singularity is at ⁠${\displaystyle z=1}$⁠. Recall the expression for the Taylor series for a function ⁠${\displaystyle g(z)}$⁠ about ⁠${\displaystyle z=a}$⁠: $${\displaystyle g(z)=g(a)+g^{\prime }(a)(z-a)+\frac{g^{″}(a)(z-a{)}^2}{2!}+\frac{g^{‴}(a)(z-a{)}^3}{3!}+\cdots .}$$ So, for ⁠${\displaystyle g(z)=\sin z}$⁠ and ⁠${\displaystyle a=1}$⁠ we have $${\displaystyle \sin z=\sin 1+(\cos 1)(z-1)+\frac{-(\sin 1)(z-1{)}^2}{2!}+\frac{-(\cos 1)(z-1{)}^3}{3!}+\cdots ,}$$ and for ⁠${\displaystyle g(z)=1/z}$⁠ and ⁠${\displaystyle a=1}$⁠ we have $${\displaystyle \frac{1}{z}=\frac{1}{(z-1)+1}=1-(z-1)+(z-1{)}^2-(z-1{)}^3+\cdots .}$$ Multiplying those two series and introducing ⁠${\displaystyle 1/(z-1)}$⁠ gives us$${\displaystyle \frac{\sin z}{z(z-1)}=\frac{\sin 1}{z-1}+(\cos 1-\sin 1)+(z-1)\left(-\frac{\sin 1}{2!}-\cos 1+\sin 1\right)+\cdots .}$$ So the residue of ⁠${\displaystyle f(z)}$⁠ at ⁠${\displaystyle z=1}$⁠ is ⁠${\displaystyle \sin 1}$⁠.

Example 3

The next example shows that, computing a residue by series expansion, a major role is played by the Lagrange inversion theorem. Let $${\displaystyle u(z):=\sum _{k\ge 1}{u}_k{z}^k}$$ be an entire function, and let $${\displaystyle v(z):=\sum _{k\ge 1}{v}_k{z}^k}$$ with positive radius of convergence, and with ⁠${\displaystyle v_1\ne 0}$⁠. So ⁠${\displaystyle v(z)}$⁠ has a local inverse ⁠${\displaystyle V(z)}$⁠ at ⁠${\displaystyle 0}$⁠, and $u(1/V(z))$ is meromorphic at ⁠${\displaystyle 0}$⁠. Then we have: $${\displaystyle {\mathrm{Res}}_0(u(1/V(z)))=\sum _{k=0}^{\mathrm{\infty }}k{u}_k{v}_k.}$$ Indeed, $${\displaystyle {\mathrm{Res}}_0(u(1/V(z)))={\mathrm{Res}}_0\left(\sum _{k\ge 1}{u}_{k}V(z{)}^{-k}\right)=\sum _{k\ge 1}{u}_k{\mathrm{Res}}_0(V(z{)}^{-k})}$$because the first series converges uniformly on any small circle around ⁠${\displaystyle 0}$⁠. Using the Lagrange inversion theorem $${\displaystyle {\mathrm{Res}}_0(V(z{)}^{-k})=k{v}_k,}$$ and we get the above expression. For example, if ${\displaystyle u(z)=z+z^2}$ and also ⁠${\displaystyle v(z)=z+z^2}$⁠, then $${\displaystyle V(z)=\frac{2z}{1+\sqrt{1+4z}}}$$ and $${\displaystyle u(1/V(z))=\frac{1+\sqrt{1+4z}}{2z}+\frac{1+2z+\sqrt{1+4z}}{2z^2}.}$$ The first term contributes ⁠${\displaystyle 1}$⁠ to the residue, and the second term contributes ⁠${\displaystyle 2}$⁠ since it is asymptotic to ⁠${\displaystyle 1/z^2+2/z}$⁠.

With the corresponding stronger symmetric assumptions on ⁠${\displaystyle u(z)}$⁠ and ⁠${\displaystyle v(z)}$⁠, it also follows that $${\displaystyle {\mathrm{Res}}_0\left(u(1/V)\right)={\mathrm{Res}}_0\left(v(1/U)\right),}$$ where ⁠${\displaystyle U(z)}$⁠ is a local inverse of ⁠${\displaystyle u(z)}$⁠ at ⁠${\displaystyle 0}$⁠.

Example 4 - Essential singularity

The function ${\displaystyle f(z)=\exp \left(\frac{5i}{z}\right)}$ is holomorphic on ${\displaystyle G=\mathbb{C}\setminus \{0\}}$. ${\displaystyle f}$ has the following Laurent series: $${\displaystyle f(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\cdot {\left(\frac{5i}{z}\right)}^n=\sum _{n=-\mathrm{\infty }}^0\underset{=a_n}{\underbrace{\frac{(5i{)}^{-n}}{(-n)!}}}\cdot z^n}$$ The residue of ${\displaystyle f}$ in ${\displaystyle z_0=0}$ is the coefficient ${\displaystyle a_{-1}}$ in the Laurent series with the center ${\displaystyle z_0=0}$ as essential singularity: $${\displaystyle Res(f,0)=a_{-1}=\frac{(5i{)}^{-(-1)}}{(-(-1))!}=5i}$$