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