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