How Do You Solve a Complex Integral Using Cauchy-Goursat's Theorem?

[Terrell]

Yes, it is late here (4 am) and I am suffering from insomnia so not completely alert ...

Yes, you can apply the Cauchy integral formula with z0=0z0=0z_0 = 0. The more general theorem is the residue theorem.

Got it! Thank you, I think I can handle this from here. Sorry if I am taking a while, I am multi-problem solving right now . Exam is tomorrow.

 

[Terrell]

Just to have closure. I'll put the entire solution here. By letting $z=e^{i\theta}$ and $dz=ie^{i\theta}d\theta$. we have
\begin{align}
\frac{1}{ie^{i\theta}}\int_{c}\cos^2(\frac{\pi}{6}+2e^{i\theta})ie^{i\theta}d\theta = \frac{1}{i}\int_{c}\frac{\cos^2(\frac{\pi}{6}+2z)}{z}dz
\end{align}
Since $\frac{\cos^2(\frac{\pi}{6}+2z)}{z}$ is analytic everywhere except at the origin - which is encapsulated by the given contour/curve we are integrating on, then by the Cauchy Integral Formula, letting $f(z)=\cos^2(\frac{\pi}{6}+2z)$ so $f(0)=3/4$ we have
\begin{align}
\int_c\frac{f(z)}{z}dz=2\pi i\frac{3}{4}=\frac{3\pi i}{2}
\end{align}

 

[Ray Vickson]

Just to have closure. I'll put the entire solution here. By letting $z=e^{i\theta}$ and $dz=ie^{i\theta}d\theta$. we have
\begin{align}
\frac{1}{ie^{i\theta}}\int_{c}\cos^2(\frac{\pi}{6}+2e^{i\theta})ie^{i\theta}d\theta = \frac{1}{i}\int_{c}\frac{\cos^2(\frac{\pi}{6}+2z)}{z}dz
\end{align}
Since $\frac{\cos^2(\frac{\pi}{6}+2z)}{z}$ is analytic everywhere except at the origin - which is encapsulated by the given contour/curve we are integrating on, then by the Cauchy Integral Formula, letting $f(z)=\cos^2(\frac{\pi}{6}+2z)$ so $f(0)=3/4$ we have
\begin{align}
\int_c\frac{f(z)}{z}dz=2\pi i\frac{3}{4}=\frac{3\pi i}{2}
\end{align}

Good, so you have used that "famous" integral I mentioned. Every single textbook about complex analysis will have lots of material using that integral formula, since it is possibly one of the most important formulas in the whole field. In applications it is used over and over again, roughly in the manner that you just now used it.