Solved Problem on Cauchy's Integral Formula
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c) \( \displaystyle \oint_{|z+1|=1}\frac{dz}{(z-1)^{3}(z+1)^{3}} \)
The path of integration is given by the circle of radius 1, centered at the point (−1, 0),
traversed counterclockwise (Figure 1).
The general form of Cauchy's Integral Formula is given by
The general form of Cauchy's Integral Formula is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{f^{(n)}(z_{0})=\frac{{n}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz} \tag{I}
\end{gather}
\]
Identifying the terms of the integral
\[
\begin{gather}
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-
\bbox[#FFD9CC,2px]
{z_{0}}
\right)^{
\bbox[#FFCC66,2px]
{n}
+1}}\;dz=\oint_{{|z+1|=1}}
\bbox[#FFFF66,2px]
{{\frac{1}{(z-1)^{3}}}}
\frac{1}{(z+
\bbox[#FFD9CC,2px]
{1}
)^{
\bbox[#FFCC66,2px]
{2}
+1}}\;dz
\end{gather}
\]
the point \( z+1=0\Rightarrow z=-1 \), lies inside the region determined by de closed contour C, will be used in the calculation of the integral we have \( f(z)=\frac{1}{(z-1)^{3}} \), z0 = −1 and n = 2, writing the expression (I)
\[
\begin{gather}
\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt]
\oint_{|z+1|=1}\frac{dz}{(z-1)^{3}(z+1)^{3}}=\frac{2\pi\mathrm{i}}{2!}\;f^{(2)}(-1)
\end{gather}
\]
Calculation of the second derivative of
\( \displaystyle f(z)=\frac{1}{(z-1)^{3}} \)
Rewriting the function f(z) as \( f(z)=(z-1)^{-3} \)
the function f(z) is a composite function using the Chain Rule
Rewriting the function f(z) as \( f(z)=(z-1)^{-3} \)
the function f(z) is a composite function using the Chain Rule
\[
\begin{gather}
\frac{du[v(z)]}{dz}=\frac{du}{dv}\frac{dv}{dz}
\end{gather}
\]
where
\( u(v)=v^{-3} \)
and
\( v(z)=(z-1) \)
\[
\begin{gather}
\frac{df}{dz}=\frac{d\left(v^{-3}\right)}{dv}\frac{d(z-1)}{dz}\\[5pt]
\frac{df}{dz}=-3v^{-3-1}(-1)\\[5pt]
\frac{df}{dz}=-3(z-1)^{-4}
\end{gather}
\]
second differentiation and applying the Chain Rule, where
\( u(v)=-3v^{-4} \)
and
\( v(z)=(z-1) \)
\[
\begin{gather}
\frac{d^{2}f}{dz^{2}}=\frac{d\left(3v^{-4}\right)}{dv}\frac{d(z-1)}{dz}\\[5pt]
\frac{d^{2}f}{dz^{2}}=-12v^{-4-1}(-1)\\[5pt]
\frac{d^{2}f}{dz^{2}}=12(z-1)^{-5}
\end{gather}
\]
\[
\begin{gather}
f^{(2)}(z)=\frac{12}{(z-1)^{5}}
\end{gather}
\]
\[
\begin{gather}
\oint_{{|z+1|=1}}\frac{dz}{(z-1)^{3}(z+1)^{3}}=\frac{\cancel{2}\pi\mathrm{i}}{\cancel{2}\times 1}\;\frac{12}{(-1-1)^{5}}\\[5pt]
\oint_{|z+1|=1}\frac{dz}{(z-1)^{3}(z+1)^{3}}=\pi\mathrm{i}\;\frac{12}{(-2)^{5}}\\[5pt]
\oint_{|z+1|=1}\frac{dz}{(z-1)^{3}(z+1)^{3}}=-\pi\mathrm{i}\;\frac{\cancelto{3}{12}}{\cancelto{8}{32}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\oint_{|z+1|=1}\frac{dz}{(z-1)^{3}(z+1)^{3}}=-\;\frac{3\pi\mathrm{i}}{8}}
\end{gather}
\]
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