Solved Problem on Cauchy's Integral Formula
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d)   \( \displaystyle \oint_{|z|=3}\frac{\cos (z^{2}+3z-1)}{(2z+3)^{2}}\;dz \)


Writing the integral as
\[ \begin{gather} \oint_{{|z|=3}}\frac{\cos(z^{2}+3z-1)}{\left[2\left(z+\dfrac{3}{2}\right)\right]^{2}}\;dz\\[5pt] \oint_{{|z|=3}}\frac{\cos(z^{2}+3z-1)}{4\left(z+\dfrac{3}{2}\right)^{2}}\;dz \end{gather} \]
The path of integration is given by the circle of radius 3, centered at the origin (o, 0), traversed counterclockwise (Figure 1).
Figure 1
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
\[ \begin{gather} \frac{{ \bbox[#FFCC66,2px] {n} }!}{2\pi \mathrm{i}}\;\oint_{{C}}\frac{ \bbox[#FFFF66,2px] {f(z)} }{\left(z- \bbox[#FFD9CC,2px] {z_{0}} \right)^{ \bbox[#FFCC66,2px] {n} +1}}\;dz=\oint_{{|z|=3}} \bbox[#FFFF66,2px] {{\frac{\cos(z^{2}+3z-1)}{4}}} \frac{1}{\left[z-\left( \bbox[#FFD9CC,2px] {{-\dfrac{3}{2}}} \right)\right]^{ \bbox[#FFCC66,2px] {1} +1}}\;dz \end{gather} \]

the point   \( z+\frac{3}{2}=0\Rightarrow z=-{\frac{3}{2}} \)   lies inside the region determined by de closed contour C, will be used in the calculation of the integral we have   \( f(z)=\frac{\cos (z^{2}+3z-1)}{4} \),   \( z_{0}=-{\frac{3}{2}} \),   and n = 1, writing the expression (I)
\[ \begin{gather} \oint_{C}\frac{f(z)}{\left(z-z_{0}\right)^{n+1}}\;dz=\frac{2\pi\mathrm{i}}{n!}\;f^{(n)}(z_{0})\\[5pt] \oint_{|z|=3}\frac{\cos (z^{2}+3z-1)}{(2z+3)^{2}}\;dz=\frac{2\pi\mathrm{i}}{1!}\;f^{(1)}\left(-{\frac{3}{2}}\right) \end{gather} \]

Calculation of the derivative of    \( \displaystyle f(z)=\frac{\cos (z^{2}+3z-1)}{4} \)

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   \( u(v)=\frac{\cos v}{4} \)   and   \( v(z)=(z^{2}+3z-1) \)
\[ \begin{gather} \frac{df}{dz}=\frac{d\left(\frac{\cos v}{4}\right)}{dv}\frac{d(z^{2}+3z-1)}{dz}\\[5pt] \frac{df}{dz}=-{\frac{\sin v}{4}}(2z+3)\\[5pt] \frac{df}{dz}=-{\frac{(2z+3)}{4}}\sin (z^{2}+3z-1) \end{gather} \]
\[ \begin{gather} f^{(1)}(z)=-{\frac{(2z+3)}{4}}\sin (z^{2}+3z-1) \end{gather} \]

\[ \begin{gather} \oint_{|z|=3}\frac{\cos(z^{2}+3z-1)}{(2z+3)^{2}}\;dz=\frac{2\pi\mathrm{i}}{1}\;\left\{-{\frac{\left[2.\left(-{\frac{3}{2}}\right)+3\right]}{4}}\sin \left[\left(-{\frac{3}{2}}\right)^{2}+3\times\left(-{\frac{3}{2}}\right)-1\right]\right\}\\[5pt] \oint_{|z|=3}\frac{\cos (z^{2}+3z-1)}{(2z+3)^{2}}\;dz=2\pi\mathrm{i}\;\left\{-{\frac{[-3+3]}{4}}\sin \left[\frac{9}{4}-\frac{9}{2}-1\right]\right\}\\[5pt] \oint_{|z|=3}\frac{\cos (z^{2}+3z-1)}{(2z+3)^{2}}\;dz=2\pi\mathrm{i}\;\left\{-{\frac{0}{4}}\times\sin \left[-{\frac{13}{4}}\right]\right\} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\oint_{{|z|=3}}\frac{\cos (z^{2}+3z-1)}{(2z+3)^{2}}\;dz=0} \end{gather} \]
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