Solved Problem on Cauchy's Integral Formula
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a) \( \displaystyle \oint_{|z+\mathrm{i}|=1}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz \)
The path of integration is given by the circle of radius 1, centered at the point (0, −1),
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
\[
\begin{gather}
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-
\bbox[#FFD9CC,2px]
{z_{0}}
\right)^{
\bbox[#FFCC66,2px]
{n}
+1}}\;dz=\oint_{{|z+\mathrm{i}|=1}}\frac{
\bbox[#FFFF66,2px]
{{\sin z}}
}{[z-(
\bbox[#FFD9CC,2px]
{{-\mathrm{i}}}
)]^{
\bbox[#FFCC66,2px]
{2}
+1}}\;dz
\end{gather}
\]
the point \( z+\mathrm{i}=0\Rightarrow z=-\mathrm{i} \), lies inside the region determined by de closed contour C, will be used in the calculation of the integral we have \( f(z)=\sin z \), z0 = −i, and n = 2, writing the expression (I)
\[
\begin{gather}
\oint_{C}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt]
\oint_{|z+\mathrm{i}|=1}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\frac{2\pi\mathrm{i}}{2!}\;f^{(2)}(-\mathrm{i})
\end{gather}
\]
Calculation of the second derivative of
\( \displaystyle f(z)=\sin z \)
\[
\begin{gather}
\frac{df}{dz}=\cos z\\[10pt]
\frac{d^{2}f}{dz^{2}}=-\sin z
\end{gather}
\]
\[
\begin{gather}
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\frac{2\pi\mathrm{i}}{2\times 1}\;[-\sin (-\mathrm{i})]
\end{gather}
\]
as sine is an odd function
\( \sin (-\theta)=-\operatorname{sen}(\theta) \)
\[
\begin{gather}
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\mathrm{i}\;[-(-\sin \;\mathrm{i})]\\[5pt]
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\mathrm{i}\;\sin \;\mathrm{i}
\end{gather}
\]
From the exponential form of the sine
\( \displaystyle \sin z=\frac{\operatorname{e}^{\mathrm{i}z}-\operatorname{e}^{-\mathrm{i}z}}{2\mathrm{i}} \)
\[
\begin{gather}
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\mathrm{i}\;\left(\frac{\operatorname{e}^{\mathrm{i}.\mathrm{i}}-\operatorname{e}^{-\mathrm{i}.\mathrm{i}}}{2\mathrm{i}}\right)\\[5pt]
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\;\left(\frac{\operatorname{e}^{\mathrm{i}^{2}}-\operatorname{e}^{-\mathrm{i}^{2}}}{2}\right)\\[5pt]
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\;\left(\frac{\operatorname{e}^{-1}-\operatorname{e}^{-(-1)}}{2}\right)\\[5pt]
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\;\left(\frac{\operatorname{e}^{-1}-\operatorname{e}^{1}}{2}\right)\\[5pt]
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=\pi\;\left(-{\frac{\operatorname{e}^{1}-\operatorname{e}^{-1}}{2}}\right)\\[5pt]
\oint_{{|z+\mathrm{i}|=1}}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=-\pi\;\left(\frac{\operatorname{e}^{1}-\operatorname{e}^{-1}}{2}\right)
\end{gather}
\]
From the exponential form of the hyperbolic sine
\( \displaystyle \sinh z=\frac{\operatorname{e}^{z}-\operatorname{e}^{-z}}{2} \)
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\oint_{|z+\mathrm{i}|=1}\frac{\sin z}{(z+\mathrm{i})^{3}}\;dz=-\pi\;\sinh 1}
\end{gather}
\]
Note: We can calculate the numerical value using e = 2.71818 e π = 3.14159
\[
\begin{split}
-\pi\;\left(\frac{\operatorname{e}^{1}-\operatorname{e}^{-1}}{2}\right) &\Rightarrow-3.14159\times\;\left(\frac{2.71818^{1}-2.71818^{-1}}{2}\right)\Rightarrow\\[5pt]
&\Rightarrow-\frac{3.14159}{2}\times\;\left(2.71818-\frac{1}{2.71818}\right)\Rightarrow\\[5pt]
&\Rightarrow-1.57079\times\;\left(\frac{2.71818.2.71818-1}{2.71818}\right)\Rightarrow\\[5pt]
&\Rightarrow-1.57079\times\;\left(\frac{7.38850-1}{2.71818}\right)\Rightarrow\\[5pt]
&\Rightarrow-1.57079\times\;\left(\frac{6.38850}{2.71818}\right)\Rightarrow\\[5pt]
&\Rightarrow-1.57079\times\;\left(\frac{6.38850}{2.71818}\right)\Rightarrow\\[5pt]
&\Rightarrow-1.57079\times\;\left(1.71817\right)\Rightarrow-2.69888
\end{split}
\]
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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 .