Solved Problem on Cauchy's Integral Formula
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d) \( \displaystyle \oint_{{|z|=1/2}}\frac{z^{2}+2z+3}{3z-1}\;dz \)
The path of integration is given by the circle of radius
\( \frac{1}{2} \),
centered at the origin (0, 0), traversed
counterclockwise (Figure 1).
The general form of Cauchy's Integral Formula given by
The general form of Cauchy's Integral Formula given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{f^{(n)}(z_{0})=\frac{{n}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz} \tag{I}
\end{gather}
\]
Identifying the terms of the integral
\[
\begin{gather}
\frac{{{n}}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{{f(z)}}{\left(z-{z_{0}}\right)^{{n}+1}}\;dz=\oint_{{|z|=1/2}}\frac{z^{2}+2z+3}{3z-1}\;dz
\end{gather}
\]
\[
\begin{gather}
\frac{n!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{{f(z)}}{\left(z-z_{0}\right)^{n+1}}\;dz=\oint_{{|z|=1/2}}\frac{{z^{2}+2z+3}}{3\left(z-{\frac{1}{3}}\right)^{0+1}}\;dz\\[5pt]
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-
\bbox[#FFD9CC,2px]
{z_{0}}
\right)^{
\bbox[#FFCC66,2px]
{n}
+1}}\;dz=\oint_{|z|=1/2}
\bbox[#FFFF66,2px]
{\frac{{z^{2}+2z+3}}{3}}
\frac{1}{\left(z-{
\bbox[#FFD9CC,2px]
{\frac{1}{3}}
}\right)^{
\bbox[#FFCC66,2px]
{0}
+1}}
\end{gather}
\]
the point
\( z-\frac{1}{3}=0\Rightarrow z=\frac{1}{3} \)
that is inside the region determined by de closed contur C, it will be used in calculation
of the integral, we have
\( f(z)=\frac{z^{2}+2z+3}{3} \),
\( z_{0}=\frac{1}{3} \)
and n = 0, writing the expression (I) for the given integral
\[
\begin{gather}
\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt]
\oint_{{|z|=1/2}}\frac{z^{2}+2z+3}{3z-1}\;dz=\frac{2\pi\mathrm{i}}{0!}\;f^{(0)}\left(\frac{1}{3}\right)\\[5pt]
\oint_{{|z|=1/2}}\frac{{z^{2}+2z+3}}{3\left(z-{\frac{1}{3}}\right)}\;dz=\frac{2\pi\mathrm{i}}{0!}\;f^{(0)}\left(\frac{1}{3}\right)\\[5pt]
\oint_{{|z|=1/2}}\frac{{z^{2}+2z+3}}{3\left(z-{\frac{1}{3}}\right)}\;dz=2\pi\mathrm{i}\;\frac{\left[\left(\frac{1}{3}\right)^{2}+2\times \frac{1}{3}+3\right]}{3}\\[5pt]
\oint_{{|z|=1/2}}\frac{{z^{2}+2z+3}}{3\left(z-{\frac{1}{3}}\right)}\;dz=2\pi\mathrm{i}\;\times\frac{1}{3}\times\left[\frac{1}{9}+\frac{2}{3}+3\right]\\[5pt]
\oint_{{|z|=1/2}}\frac{{z^{2}+2z+3}}{3\left(z-{\frac{1}{3}}\right)}\;dz=2\pi\mathrm{i}\;\times\frac{1}{3}\times\left[\frac{1+6+27}{9}\right]\\[5pt]
\oint_{{|z|=1/2}}\frac{{z^{2}+2z+3}}{3\left(z-{\frac{1}{3}}\right)}\;dz=2\times\frac{1}{3}\times\frac{34}{3}\pi\mathrm{i}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\oint_{{|z|=1/2}}\frac{z^{2}+2z+3}{3z-1}\;dz=\frac{68}{27}\pi\mathrm{i}}
\end{gather}
\]
Note 1: The path traversed
\( |\;z\;|=1/2 \)
is a circle. For a complex number
\( z=x+\mathrm{i}y \),
the absolute value is given by
\( \sqrt{x^{2}+y^{2}\;}=1/2 \),
squaring both sides of equation
\( \left(\sqrt{x^{2}+y^{2}\;}\right)^{2}=(1/2)^{2} \),
we obtain the equation of a circle
\( (x-0)^{2}+(y-0)^{2}=(1/2)^{2}, \)
Note 2: f(0) represents the calculation of the function at the point z0 without derivative.
\[ (x-0)^{2}+(y-0)^{2}=(1/2)^{2} \]
with a radius equal to 1/2 and center at the origin (0, 0).
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 .