# calculation of contour integral

We will determine the important complex integral

 $I\;:=\;\oint_{C}\!(z\!-\!z_{0})^{n}\,dz$

where $C$ is the circumference of the circle  $|z\!-\!z_{0}|=\varrho$  taken anticlockwise and $n$ an arbitrary integer.

Let’s take the “direction angle” of the radius of $C$ as the parametre $t$, i.e.

 $t\;:=\;\arg|z\!-\!z_{0}|.$

Then on $C$, we have

 $z\!-\!z_{0}\;=\;\varrho e^{it},\quad 0\;\leqq\;t\leqq 2\pi$

and

 $dz\;=\;i\varrho e^{it}\,dt,\quad(z\!-\!z_{0})^{n}\;=\;\varrho^{n}e^{int},$

whence

 $I\;=\;\int_{0}^{2\pi}\!\varrho^{n}e^{int}i\varrho e^{it}\,dt\;=\;i\varrho^{n+1% }\!\int_{0}^{2\pi}\!e^{i(n+1)t}\,dt.$

In the case  $n=-1$  one gets trivially  $I=2i\pi$.  If  $n\neq-1$,  we obtain

 $I\;=\;i\varrho^{n+1}\!\!\operatornamewithlimits{\Big{/}}_{\!\!\!t=0}^{\,\quad 2% \pi}\!\frac{e^{i(n+1)t}}{i(n\!+\!1)}\;=\;\frac{\varrho^{n+1}}{n\!+\!1}(1\!-\!1% )\;=\;0,$

using the fact that $2i\pi$ is a period of the exponential function (http://planetmath.org/PeriodicityOfExponentialFunction).

Hence we can write the result

 $\displaystyle\oint_{C}(z\!-\!z_{0})^{n}\,dz\;=\;\begin{cases}2i\pi\quad\mbox{% if}\;\;n\;=\;-1,\\ 0\qquad\mbox{if}\;\;n\;\in\;\mathbb{Z}\!\smallsetminus\!\{-1\}.\\ \end{cases}$
Title calculation of contour integral CalculationOfContourIntegral 2013-03-22 19:14:16 2013-03-22 19:14:16 pahio (2872) pahio (2872) 8 pahio (2872) Example msc 30E20 msc 30A99 AntiderivativeOfComplexFunction SubstitutionNotation ProofOfCauchyIntegralFormula