Complex integration
Contents
Residue calculus
© The scientific sentence. 2010
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Complex integration :
Residue calculus
1. How to obtain a residue
•
The point zo is a pole of order m for f, if the
Laurent expansion of f around zo starts from j = - m, so that
a- m ≠ 0. That is
\[ \large\bf\color{brown}{
\textit f(\textit z) = \sum_{j = - m}^\infty
\textit a_j(\textit z - \textit z_o)^j}
=
\\
\]
\[ \bf\color{indigo}{
\frac{a_{- m}}{(\textit z - \textit z_o)^m} +
... + a_0 + a_1(\textit z - \textit z_o) +
a_2(\textit z - \textit z_o)^2 + ...
\\
with \; a_{- m} \ne 0
}
\]
⇒
(z - zo)m f(z) = a- m +
a- m + 1(z - zo) + ... +
a- 1(z - zo)m - 1 +
a0(z - zo)m +
a1(z - zo)m + 1 +
a2(z - zo)m + 2 +
... +
am(z - zo)2m +
...
Let's take the derivative of order m - 1 . We find:
dm - 1[(z - zo)m f(z)]dzm - 1 = a- m +
a- m + 1(z - zo) + ... +
a- 1(z - zo)m - 1 +
a0(z - zo)m +
a1(z - zo)m + 1 +
a2(z - zo)m + 2 +
... +
am(z - zo)2m +
... =
\[ \large\bf\color{brown}{
\frac{d^{m-1}}{dz^{m-1}}[(z - z_o)^m f(z)] =
(m - 1)! \; a_{-1} +
\\
\sum_{n = 1}^\infty
\textit b_n(\textit z - \textit z_o)^n
}
\]
Let's take the limit z → zo. We obtain:
\[ \large\bf\color{blue}{
\text{R(zo) = a_1 = }
}
\]
\[ \large\bf\color{blue}{
\lim_{\textit z \to zo} {
\frac{1}{(m - 1)!} \frac{d^{m-1}}{dz^{m-1}}[(z - z_o)^m f(z)]} =
}
\]
\[ \bf\color{red}{
\textit{Residue of f at z = zo.}
}
\]
2. Case of a simple pole
If zo is a simple pole of f(z),that is m = 1, we
will have:
\[ \large\bf\color{blue}{
\text{R(zo) = }
\lim_{\textit z \to zo}
[(z - z_o) f(z)] =
}
\]
\[ \bf\color{red}{
\textit {Residue of f at z = zo.}
}
\]
f(z) has a pole of order 1, so f(z) can be written as:
\[ \large\bf\color{blue}{
f(z) = \frac {g(z)}{h(z)}
}
\]
Where g(z) is analytic and non-zero at zo; and
h(zo) = 0.
Therefore
\[ \bf\color{teal}{
\text{R(zo) = }
\lim_{\textit z \to zo}
[(z - z_o) f(z)] =
\\
\lim_{\textit z \to zo} [(z - z_o) \frac {g(z)}{h(z)}]
\\
= g(zo)
\lim_{\textit z \to zo} \frac {(z - z_o)}{h(z)}
\\
= g(zo)
\lim_{\textit z \to zo} \frac {1}{\frac{h(z) - h(zo)}{z - zo}}
\\
= g(zo)
\lim_{\textit z \to zo} \frac {1}{h'(z)} =
\frac {g(zo)}{h'(zo)}
}
\]
3. Example
f(z) has two simple poles of order m = 2. zo = + i, and zo = - i.
\[ \bf\color{blue}{
\text{Res(- i) = }
\lim_{\textit z \to - i}
\frac{d}{dz}[(z + i)^2 f(z)] =
\\
\lim_{\textit z \to - i}
\frac{d}{dz}[(z + i)^2 \frac {exp \; \textit {iz} }{(\textit z + i)^2 (\textit z - i)^2} \;] =
\\
\lim_{\textit z \to - i}
\frac{d}{dz}[ \frac {exp \; \textit {iz} }{ (\textit z - i)^2}]
= 0
}
\]
\[ \bf\color{indigo}{
\text{Res(+ i) = }
\lim_{\textit z \to i}
\frac{d}{dz}[(z - i)^2 f(z)] =
\\
\lim_{\textit z \to - i}
\frac{d}{dz}[(z - i)^2 \frac {exp \; \textit {iz} }{(\textit z + i)^2 (\textit z - i)^2} \;] =
\\
\lim_{\textit z \to i}
\frac{d}{dz}[ \frac {exp \; \textit {iz} }{ (\textit z + i)^2}]
= - \frac{i}{2e}
}
\]
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