Solved Problem on Cauchy's Integral Formula
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a) \( \displaystyle \oint_{{C}}\frac{z}{(z+1)(z-2)}\;dz \)
The path is given by contour C (Figure 1), followed in a clockwise direction. The points
\( z+1=0\Rightarrow z=-1 \)
and
\( z-2=0\Rightarrow z=2 \)
are inside the region determined by the contour, they will be used in the calculation of the
integral.
To calculate the integral, we use the method of the Partial Fractions Decomposition
To calculate the integral, we use the method of the Partial Fractions Decomposition
\[
\begin{gather}
\oint_{{C}}\frac{z}{(z+1)(z-2)}\;dz \tag{I}
\end{gather}
\]
Decomposition of \( \displaystyle \frac{1}{(z+1)(z-2)} \)
\[
\begin{gather}
\frac{1}{(z+1)(z-2)}=\frac{A}{z+1}+\frac{B}{z-2}\\[5pt]
\frac{1}{(z+1)(z-2)}=\frac{A(z-2)+B(z+1)}{(z+1)(z-2)}\\[5pt]
1=A(z-2)+B(z+1) \tag{II}
\end{gather}
\]
letting
\( z-2=0\Rightarrow z=2 \)
and substituting into expression (II)
\[
\begin{gather}
1=A(2-2)+B(2+1)\\[5pt]
3B+0=1\\[5pt]
B=\frac{1}{3}
\end{gather}
\]
letting
\( z+1=0\Rightarrow z=-1 \)
and substituting into expression (II)
\[
\begin{gather}
1=A(-1-2)+B(-1+1)\\[5pt]
-3A+0=1\\[5pt]
A=-{\frac{1}{3}}
\end{gather}
\]
substituting the values of A and B
\[
\begin{gather}
\frac{1}{(z+1)(z-2)}=\frac{-{\frac{1}{3}}}{z+1}+\frac{\frac{1}{3}}{z-2}\\[5pt]
\frac{1}{(z+1)(z-2)}=-{\frac{1}{3(z+1)}}+\frac{1}{3(z-2)}
\end{gather}
\]
The integral (I) is rewritten as
\[
\begin{gather}
\oint_{{C}}\frac{z}{(z+1)(z-2)}\;dz=\underbrace{\oint_{{C}}-{\frac{z}{3(z+1)}}\;dz}_{I_{1}}+\underbrace{\oint_{{C}}\frac{z}{3(z-2)}\;dz}_{I_{2}}
\end{gather}
\]
The integral, will be given by the sum of the integrals I1 and I2.The Cauchy Integral Formula in the general form is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{f^{(n)}(z_{0})=\frac{{n}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz} \tag{III}
\end{gather}
\]
Identifying the terms of the integrals.
- Integral I1:
\[
\begin{gather}
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-
\bbox[#FFD9CC,2px]
{z_{0}}
\right)^{
\bbox[#FFCC66,2px]
{n}
+1}}\;dz=\oint_{C}
\bbox[#FFFF66,2px]
{-{\frac{z}{3}}}
\frac{1}{\left[z-
\bbox[#FFD9CC,2px]
{(-1)}
\right]^{
\bbox[#FFCC66,2px]
{0}
+1}}\;dz
\end{gather}
\]
we have
\( f(z)=-{\frac{z}{3}} \),
z0 = −1 and n = 0, writing expression (III) for the integral
\[
\begin{gather}
I_{1}=\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt]
I_{1}=\oint_{{C}}-{\frac{z}{3(z+1)}}\;dz=\frac{2\pi\mathrm{i}}{0!}\;f^{(0)}(-1)\\[5pt]
I_{1}=2\pi\mathrm{i}\;\times\left[-{\frac{(-1)}{3}}\right]\\[5pt]
I_{1}=\frac{2\pi\mathrm{i}}{3} \tag{IV}
\end{gather}
\]
- Integral I2:
\[
\begin{gather}
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-
\bbox[#FFD9CC,2px]
{z_{0}}
\right)^{
\bbox[#FFCC66,2px]
{n}
+1}}\;dz=\oint_{C}
\bbox[#FFFF66,2px]
{\frac{z}{3}}
\frac{1}{\left[z-
\bbox[#FFD9CC,2px]
{2}
\right]^{
\bbox[#FFCC66,2px]
{0}
+1}}\;dz
\end{gather}
\]
we have
\( f(z)=\frac{z}{3} \),
z0 = 2 and n = 0, writing expression (III) for the integral
\[
\begin{gather}
I_{2}=\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt]
I_{2}=\oint_{{C}}\frac{z}{3(z-2)}\;dz=\frac{2\pi\mathrm{i}}{0!}\;f^{(0)}(2)\\[5pt]
I_{2}=2\pi\mathrm{i}\;\times\frac{2}{3}\\[5pt]
I_{2}=\frac{4\pi \mathrm{i}}{3} \tag{V}
\end{gather}
\]
The result of the integral will be given by the sum of the values of (IV) and (V)
\[
\begin{gather}
\oint_{{C}}\frac{z}{(z+1)(z-2)}\;dz=I_{1}+I_{2}\\[5pt]
\oint_{{C}}\frac{z}{(z+1)(z-2)}\;dz=\frac{2\pi\mathrm{i}}{3}+\frac{4\pi \mathrm{i}}{3}\\[5pt]
\oint_{{C}}\frac{z}{(z+1)(z-2)}\;dz=\frac{6\pi\mathrm{i}}{3}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=2\pi \mathrm{i}}
\end{gather}
\]
Note 1: We do not need to know the equation that describes the contour C for the
calculation, it is enough to know if the singularity points are inside or outside the region determined
by the contour.
Note 2: f(0) represents the calculation of the function at the point z0 without derivative.
Note 2: f(0) represents the calculation of the function at the point z0 without derivative.
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .