Solved Problem on Cauchy's Integral Formula
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e) \( \displaystyle \oint_{{|z-\mathrm{i}|=2}}\frac{\sin z}{z-\frac{\pi}{4}}\;dz \)
The path of integration is given by the circle of radius 2, centered at point (0, 1), 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
\[
\begin{gather}
\frac{{
\bbox[#FFCC66,2px]
{n}
}!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{
\bbox[#FFFF66,2px]
{f(z)}
}{\left(z-
\bbox[#FFD9CC,2px]
{z_{0}}
\right)^{
\bbox[#FFCC66,2px]
{n}
+1}}\;dz=\oint_{{|z|=2}}\frac{
\bbox[#FFFF66,2px]
{\sin z}
}{\left(z-
\bbox[#FFD9CC,2px]
{\frac{\pi}{4}}
\right)^{
\bbox[#FFCC66,2px]
{0}
+1}}\;dz
\end{gather}
\]
the point
\( z-\frac{\pi}{4}=0\Rightarrow z=\frac{\pi}{4} \)
that is inside the region determined by de closed contur C, it will be used in calculation
of the integral, we have
\( f(z)=\sin z \),
\( z_{0}=\frac{\pi}{4} \)
and n = 0, writing the expression (I) for the given integral
\[
\begin{gather}
\oint_{{C}}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt]
\oint_{{|z-\mathrm{i}|=2}}\frac{\sin z}{z-\frac{\pi}{4}}\;dz=\frac{2\pi \mathrm{i}}{0!}\;f^{(0)}\left(\frac{\pi}{4}\right)\\[5pt]
\oint_{{|z-\mathrm{i}|=2}}\frac{\sin z}{z-\frac{\pi}{4}}\;dz=2\pi \mathrm{i}\;\sin \frac{\pi }{4}\\[5pt]
\oint_{{|z-\mathrm{i}|=2}}\frac{\sin z}{z-\frac{\pi}{4}}\;dz=\cancel{2}\pi \mathrm{i}\;\times \frac{\sqrt{2\;}}{\cancel{2}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\oint_{{|z-\text{i}|=2}}\frac{\sin z}{z-\frac{\pi}{4}}\;dz=\sqrt{2\;}\pi \mathrm{i}}
\end{gather}
\]
Note 1: The path traversed
\( |\;z-\mathrm{i}\;|=2 \)
is a circle. For a complex number
\( z=x+\mathrm{i}y-\mathrm{i}\Rightarrow x+\mathrm{i}(y-1), \)
Note 2: f(0) represents the calculation of the function at the point z0 without derivative.
\[ z=x+\mathrm{i}y-\mathrm{i}\Rightarrow x+\mathrm{i}(y-1) \]
the absolute value is given by
\( \sqrt{x^{2}+(y-1)^{2}\;}=2, \)
\[ \sqrt{x^{2}+(y-1)^{2}\;}=2 \]
squaring both sides of equation
\( \left(\sqrt{x^{2}+(y-1)^{2}\;}\right)^{2}=2^{2}, \)
\[ \left(\sqrt{x^{2}+(y-1)^{2}\;}\right)^{2}=2^{2} \]
we obtain the equation of a circle
\( (x-0)^{2}+(y-1)^{2}=2^{2}, \)
\[ (x-0)^{2}+(y-1)^{2}=2^{2} \]
with a radius equal to 2 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 .