Solved Problem on Cauchy's Integral Formula
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b) \( \displaystyle \oint_{|z-1|=1}\frac{dz}{(z-1)^{3}(z+1)^{3}} \)
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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