Partial fractions in complex analysis

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 ${\textstyle {\frac {1}{(az+b)^{k}}}+p(z)}$, where ${\displaystyle a}$ and ${\displaystyle b}$ are complex, ${\displaystyle k}$ is an integer, and ${\displaystyle p(z)}$ is a polynomial. Just as polynomial factorization can be generalized to the Weierstrass factorization theorem, there is an analogy to partial fraction expansions for certain meromorphic functions.

A proper rational function (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 ${\displaystyle f(z)}$ for which ${\displaystyle |f(z)|}$ goes to 0 as ${\displaystyle z}$ goes to infinity at least as quickly as ${\textstyle |{\frac {1}{z}}|}$ has an expansion with no polynomial terms.

Summarize Timeline Top Qs Fact Check

Let ${\displaystyle f(z)}$ be a function meromorphic in the finite complex plane with poles at ${\displaystyle \lambda _{1},\lambda _{2},...}$ and let ${\displaystyle (\Gamma _{1},\Gamma _{2},...)}$ be a sequence of simple closed curves such that:

• The origin lies inside each curve ${\displaystyle \Gamma _{k}}$
• No curve passes through a pole of ${\displaystyle f}$
• ${\displaystyle \Gamma _{k}}$ lies inside ${\displaystyle \Gamma _{k+1}}$ for all ${\displaystyle k}$
• ${\displaystyle \lim _{k\rightarrow \infty }d(\Gamma _{k})=\infty }$, where ${\displaystyle d(\Gamma _{k})}$ gives the distance from the curve to the origin
• one more condition of compatibility with the poles ${\displaystyle \lambda _{k}}$, described at the end of this section

Suppose also that there exists an integer ${\displaystyle p}$ such that

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

Writing ${\displaystyle \operatorname {PP} (f(z);z=\lambda _{k})}$ for the principal part of the Laurent expansion of ${\displaystyle f}$ about the point ${\displaystyle \lambda _{k}}$, we have

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

if ${\displaystyle p=-1}$. If ${\displaystyle p>-1}$, then

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

where the coefficients ${\displaystyle c_{j,k}}$ are given by

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

${\displaystyle \lambda _{0}}$ should be set to 0, because even if ${\displaystyle f(z)}$ itself does not have a pole at 0, the residues of ${\textstyle {\frac {f(z)}{z^{j+1}}}}$ at ${\displaystyle z=0}$ must still be included in the sum.

Note that in the case of ${\displaystyle \lambda _{0}=0}$, we can use the Laurent expansion of ${\displaystyle f(z)}$ 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 contributed are exactly the regular part of the Laurent series up to ${\displaystyle z^{p}}$.

For the other poles ${\displaystyle \lambda _{k}}$ where ${\displaystyle k\geq 1}$, ${\textstyle {\frac {1}{z^{j+1}}}}$ 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 ${\displaystyle \lambda _{k}}$ is inside ${\displaystyle \Gamma _{n}}$, then ${\displaystyle \lambda _{j}}$ is also inside ${\displaystyle \Gamma _{n}}$ for all ${\displaystyle j<k}$.

Summarize Timeline Top Qs Fact Check

The simplest meromorphic functions with an infinite number of poles are the non-entire trigonometric functions. As an example, ${\displaystyle \tan(z)}$ is meromorphic with poles at ${\textstyle (n+{\frac {1}{2}})\pi }$, ${\displaystyle n=0,\pm 1,\pm 2,...}$ The contours ${\displaystyle \Gamma _{k}}$ will be squares with vertices at ${\displaystyle \pm \pi k\pm \pi ki}$ traversed counterclockwise, ${\displaystyle k>1}$, which are easily seen to satisfy the necessary conditions.

On the horizontal sides of ${\displaystyle \Gamma _{k}}$,

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

so

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

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

${\displaystyle \sinh(x)<\cosh(x)}$ for all real ${\displaystyle x}$, 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 ${\displaystyle x>0}$, ${\displaystyle \coth(x)}$ is continuous, decreasing, and bounded below by 1, so it follows that on the horizontal sides of ${\displaystyle \Gamma _{k}}$, ${\displaystyle |\tan(z)|<\coth(\pi )}$. Similarly, it can be shown that ${\displaystyle |\tan(z)|<1}$ on the vertical sides of ${\displaystyle \Gamma _{k}}$.

With this bound on ${\displaystyle |\tan(z)|}$ we can see that

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

That is, the maximum of ${\textstyle |{\frac {1}{z}}|}$ on ${\displaystyle \Gamma _{k}}$ occurs at the minimum of ${\displaystyle |z|}$, which is ${\displaystyle k\pi }$.

Therefore ${\displaystyle p=0}$, and the partial fraction expansion of ${\displaystyle \tan(z)}$ 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 ${\displaystyle \tan(z)}$ are simple and have residue -1:

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

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

We can ignore ${\displaystyle \lambda _{0}=0}$, since both ${\displaystyle \tan(z)}$ and ${\textstyle {\frac {\tan(z)}{z}}}$ are analytic at 0, so there is no contribution to the sum, and ordering the poles ${\displaystyle \lambda _{k}}$ so that ${\textstyle \lambda _{1}={\frac {\pi }{2}},\lambda _{2}={\frac {-\pi }{2}},\lambda _{3}={\frac {3\pi }{2}}}$, etc., gives

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

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