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.
\[
\begin{gather}
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz \tag{I}
\end{gather}
\]
The integral (I) is rewritten as a sum of two integrals calculated over the contours
C1, calculated at the point z0 = 2, and over the contour
C2 calculated at the point z0 = −1
\[
\begin{gather}
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=\oint_{{C_{1}=2}}\frac{z}{(z+1)}\frac{1}{(z-2)}\;dz+\oint_{{C_{2}=-1}}\frac{z}{(z-2)}\frac{1}{(z+1)}\;dz
\end{gather}
\]
The Cauchy Integral Formula in the general form is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{f^{(n)}(z_{0})=\frac{{n}!}{2\pi \mathrm{i}}\;\oint_{C}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz} \tag{II}
\end{gather}
\]
Identifying the terms of the integrals.
\[
\begin{gather}
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi\mathrm{i}}\;\oint_{C}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-{
\bbox[#FFD9CC,2px]
{z_{0}}
}\right)^{{
\bbox[#FFCC66,2px]
{n}
}+1}}\;dz
=\oint_{{C_{1}=2}}
\bbox[#FFFF66,2px]
{{\frac{z}{(z+1)}}}
\frac{1}{(z-
\bbox[#FFD9CC,2px]
{2}
)^{
\bbox[#FFCC66,2px]
{0}
+1}}\;dz+\oint_{{C_{2}=-1}}
\bbox[#FFFF66,2px]
{{\frac{z}{(z-2)}}}
\frac{1}{(z+
\bbox[#FFD9CC,2px]
{1}
)^{
\bbox[#FFCC66,2px]
{0}
+1}}\;dz
\end{gather}
\]
we have
\( f_{1}(z)=\frac{z}{z+1} \),
z01 = 2 e n = 0 and
\( f_{2}(z)=\frac{z}{z-2} \),
z02 = −1 e n = 0, writing the expression (III) for each integrals
\[
\begin{gather}
\oint_{C}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f_{1}^{(n)}(z_{01})+\frac{2\pi\mathrm{i}}{n!}\;f_{2}^{(n)}(z_{02})\\[5pt]
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=\frac{2\pi\mathrm{i}}{0!}\;f_{1}^{(0)}(2)+\frac{2\pi\mathrm{i}}{0!}\;f_{2}^{(0)}(-1)\\[5pt]
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=\frac{2\pi\mathrm{i}}{0!}\;\left(\frac{2}{2+1}\right)+\frac{2\pi\mathrm{i}}{0!}\;\left(\frac{-1}{-1-2}\right)\\[5pt]
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=2\pi\mathrm{i}\;\left(\frac{2}{3}\right)+2\pi\mathrm{i}\;\left(\frac{-1}{-3}\right)\\[5pt]
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=\frac{4\pi \mathrm{i}}{3}+\frac{2\pi\mathrm{i}}{3}\\[5pt]
\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=\frac{6\pi\mathrm{i}}{3}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\oint_{C}\frac{z}{(z+1)(z-2)}\;dz=2\pi \mathrm{i}}
\end{gather}
\]
Observação 1: Não precisamos conhecer a equação que descreve o contorno C para o
cálculo, basta saber se os pontos de singularidade estão dentro ou fora da região determinada pelo
contorno.
Observação 2: f(0) representa o cálculo da função no ponto z0 sem derivada.
Observação 2: f(0) representa o cálculo da função no ponto z0 sem derivada.
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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 .