Partial fractions in complex analysis

In complex analysis, a partial fraction expansion is a way of writing a meromorphic function f(z) as an infinite sum of rational functions and polynomials. When f(z) is a rational function, this reduces to the usual method of partial fractions.

Motivation

By using polynomial long division and the partial fraction technique from algebra, any rational function can be written as a sum of terms of the form 1 / (az + b)^k + p(z), where a and b are complex, k is an integer, and p(z) is a polynomial . Just as polynomial factorization can be generalized to the Weierstrass factorization theorem, there is an analogy to partial fraction expansions for arbitrary meromorphic functions.

A proper rational function, i.e. one for which the degree of the denominator is greater than the degree of the numerator, has a partial fraction expansion with no polynomial terms. Similarly, a meromorphic function f(z) for which |f(z)| goes to 0 as z goes to infinity at least as quickly as |1/z|, has an expansion with no polynomial terms.

Calculation

Let f(z) be a function meromorphic in the finite complex plane with poles at λ₁, λ₂, ..., and let (Γ₁, Γ₂, ...) be a sequence of simple closed curves such that:

• The origin lies inside each curve Γ_k
• No curve passes through a pole of f
• Γ_k lies inside Γ_k+1 for all k
• ${\displaystyle \lim _{k\rightarrow \infty }d(\Gamma _{k})=\infty }$ , where d(Γ_k) gives the distance from the curve to the origin

Suppose also that there exists an integer p such that

${\displaystyle \lim _{k\rightarrow \infty }\oint _{\Gamma _{k}}\left|{\frac {f(z)}{z^{p+1}}}\right|dz<\infty }$ 

Writing PP(f(z); z = λ_k) for the principal part of the Laurent expansion of f about the point λ_k, we have

${\displaystyle f(z)=\sum _{k=0}^{\infty }\operatorname {PP} (f(z);z=\lambda _{k}),}$ 

if p = -1, and if p > -1,

${\displaystyle f(z)=\sum _{k=0}^{\infty }(\operatorname {PP} (f(z);z=\lambda _{k})+c_{0,k}+c_{1,k}z+\cdots +c_{p,k}z^{p}),}$ 

where the coefficients c_j,k are given by

${\displaystyle c_{j,k}=\operatorname {Res} _{z=\lambda _{k}}{\frac {f(z)}{z^{j+1}}}}$ 

Note that if λ_k = 0, we can use the Laurent expansion of f(z) about the origin to get

${\displaystyle f(z)={\frac {a_{-m}}{z^{m}}}+{\frac {a_{-m+1}}{z^{m-1}}}+\cdots +a_{0}+a_{1}z+\cdots }$ 

${\displaystyle c_{j,k}=\operatorname {Res} _{z=0}\left({\frac {a_{-m}}{z^{m+j+1}}}+{\frac {a_{-m+1}}{z^{m+j}}}+\cdots +{\frac {a_{j}}{z}}+\cdots \right)=a_{j},}$ 

${\displaystyle \sum _{j=0}^{p}c_{j,k}z^{j}=a_{0}+a_{1}z+\cdots +a_{p}z^{p}}$ 

so that the polynomial terms in the partial fraction expansion are exactly the regular part of the Laurent series up to z^p.

If λ_k is not 0, 1/z^j+1 can be pulled out of the residue calculations:

${\displaystyle c_{j,k}={\frac {1}{\lambda _{k}^{j+1}}}\operatorname {Res} _{z=\lambda _{k}}f(z)}$ 

${\displaystyle \sum _{j=0}^{p}c_{j,k}z^{j}=[\operatorname {Res} _{z=\lambda _{k}}f(z)]\sum _{j=0}^{p}{\frac {1}{\lambda _{k}^{j+1}}}z^{j}}$ 

To avoid issues with convergence, the poles should be ordered so that if λ_k is inside Γ_n, then λ_j is also inside Γ_n for all j < k.

Example

File:ContourSquares.PNG

The simplest examples of meromorphic functions with an infinite number of poles are the non-entire trigonometric functions, so take the function tan(z). tan(z) is meromorphic with poles at (n + 1/2)π, n = 0, ±1, ±2, ... The contours Γ_k will be squares with vertices at ±πk ± πki traversed counterclockwise, k > 1, which are easily seen to satisfy the necessary conditions.

On the horizontal sides of Γ_k,

${\displaystyle z=t\pm \pi ki,\ \ t\in [-\pi k,\pi k],}$ 

so

${\displaystyle |\tan(z)|^{2}=\left|{\frac {\sin(t)\cosh(\pi k)\pm i\cos(t)\sinh(\pi k)}{\cos(t)\cosh(\pi k)\pm i\sin(t)\sinh(\pi k)}}\right|^{2}}$ 

${\displaystyle |\tan(z)|^{2}={\frac {\sin ^{2}(t)\cosh ^{2}(\pi k)+\cos ^{2}(t)\sinh ^{2}(\pi k)}{\cos ^{2}(t)\cosh ^{2}(\pi k)+\sin ^{2}(t)\sinh ^{2}(\pi k)}}}$ 

sinh(x) < cosh(x) for all real x, which yields

${\displaystyle |\tan(z)|^{2}<{\frac {\cosh ^{2}(\pi k)(\sin ^{2}(t)+\cos ^{2}(t))}{\sinh ^{2}(\pi k)(\cos ^{2}(t)+\sin ^{2}(t))}}=\coth ^{2}(\pi k)}$ 

For x > 0, coth(x) is continuous, decreasing, and bounded below by 1, so it follows that on the horizontal sides of Γ_k, |tan(z)| < coth(π). Similarly, it can be shown that |tan(z)| < 1 on the vertical sides of Γ_k.

With this bound on |tan(z)| we can see that

${\displaystyle \oint _{\Gamma _{k}}\left|{\frac {\tan(z)}{z}}\right|dz\leq \operatorname {length} (\Gamma _{k})\max _{z\in \Gamma _{k}}\left|{\frac {\tan(z)}{z}}\right|<8k\pi {\frac {\coth(\pi )}{k\pi }}=8\coth(\pi )<\infty .}$ 

(The maximum of |1/z| on Γ_k occurs at the minimum of |z|, which is kπ).

Therefore p = 0, and the partial fraction expansion of tan(z) looks like

${\displaystyle \tan(z)=\sum _{k=0}^{\infty }(\operatorname {PP} (\tan(z);z=\lambda _{k})+\operatorname {Res} _{z=\lambda _{k}}{\frac {\tan(z)}{z}}).}$ 

The principal parts and residues are easy enough to calculate, as all the poles of tan(z) are simple and have residue -1:

${\displaystyle \operatorname {PP} (\tan(z);z=(n+{\frac {1}{2}})\pi )={\frac {-1}{z-(n+{\frac {1}{2}})\pi }}}$ 

${\displaystyle \operatorname {Res} _{z=(n+{\frac {1}{2}})\pi }{\frac {\tan(z)}{z}}={\frac {-1}{(n+{\frac {1}{2}})\pi }}}$ 

Ordering the poles λ_k so that λ₁ = π/2, λ₂ = -π/2, λ₃ = 3π/2, etc., gives

${\displaystyle \tan(z)=\sum _{k=0}^{\infty }\left[\left({\frac {-1}{z-(k+{\frac {1}{2}})\pi }}-{\frac {1}{(k+{\frac {1}{2}})\pi }}\right)+\left({\frac {-1}{z+(k+{\frac {1}{2}})\pi }}+{\frac {1}{(k+{\frac {1}{2}})\pi }}\right)\right]}$ 

${\displaystyle \tan(z)=\sum _{k=0}^{\infty }{\frac {-2z}{z^{2}-(k+{\frac {1}{2}})^{2}\pi ^{2}}}}$ 

Application to infinite products

Because the partial fraction expansion often yields sums of 1/(a+bz), it can be useful in finding a way to write a function as an infinite product; integrating both sides gives a sum of logarithms, and exponentiating gives the desired product:

${\displaystyle \tan(z)=-\sum _{k=0}^{\infty }\left({\frac {1}{z-(k+{\frac {1}{2}})\pi }}+{\frac {1}{z+(k+{\frac {1}{2}})\pi }}\right)}$ 

${\displaystyle \int _{0}^{z}\tan(w)dw=\log \sec z}$ 

${\displaystyle \int _{0}^{z}{\frac {1}{w\pm (k+{\frac {1}{2}})\pi }}dw=\log \left(1\pm {\frac {z}{(k+{\frac {1}{2}})\pi }}\right)}$ 

Applying some logarithm rules,

${\displaystyle \log \sec z=-\sum _{k=0}^{\infty }\left(\log \left(1-{\frac {z}{(k+{\frac {1}{2}})\pi }}\right)+\log \left(1+{\frac {z}{(k+{\frac {1}{2}})\pi }}\right)\right)}$ 

${\displaystyle \log \cos z=\sum _{k=0}^{\infty }\log \left(1-{\frac {z^{2}}{(k+{\frac {1}{2}})^{2}\pi ^{2}}}\right),}$ 

which finally gives

${\displaystyle \cos z=\prod _{k=0}^{\infty }\left(1-{\frac {z^{2}}{(k+{\frac {1}{2}})^{2}\pi ^{2}}}\right).}$ 

See also

• Partial fractions
• Path integral

References

• Markushevich, A.I. Theory of functions of a complex variable. Trans. Richard A. Silverman. Vol. 2. Englewood Cliffs, N.J.: Prentice-Hall, 1965.