Cauchys integral formula

Question

I have been tasked with finding $\int_\gamma \frac{Re(z)}{2z-1}$ where $\gamma$ is the unit circle. I've tried this a few times and each time I get a different answer, I've ended up with $i\pi/2, -\pi/2 $ and $i\pi/4$. Not sure where I'm going wrong/if I'm even going right! After this, I was told to use the substitution $z=e^{it}$ to solve $\int_0^{2\pi} \frac{cos^2t}{5-4sint}dt$ I have performed the substitution and have no idea where to go from there. Any help will be appreciated!

Answer 1

If you want to use contour integration, note that $2\text{Re}(z)=z+1/z$ on the unit circle. So you get $$\frac12\int_\gamma\frac{z+z^{-1}}{2z-1}\,dz$$ which involves looking at two simple poles.

Answer 2

A no-brainer substitution approach leads to

$$I=\oint_{\|z\|=1}\frac{\text{Re}(z)}{2z-1}\,dz = i\int_{0}^{2\pi}\frac{\cos\theta}{2e^{i\theta}-1}e^{i\theta}\,d\theta=i\int_{0}^{2\pi}\frac{\cos(\theta)\,d\theta}{2-e^{-i\theta}}$$ then, by geometric series, to: $$ I = i\sum_{n\geq 0}\frac{1}{2^{n+1}}\int_{0}^{2\pi}\cos(\theta)e^{-ni\theta}\,d\theta$$ where the orthogonality relations give that only the contribute associated with $n=1$ differs from zero. It follows that $$ I = \frac{i}{4}\int_{0}^{2\pi}\cos^2(\theta)\,d\theta = \frac{\pi i}{4}.$$ About the second part, you don't really need residues. $$ \int_{0}^{2\pi}\frac{\cos^2(\theta)d\theta}{5-4\sin\theta}=\int_{-\pi}^{\pi}\frac{\cos^2(\theta)d\theta}{5+4\sin\theta}=10\int_{0}^{\pi}\frac{\cos^2(\theta)d\theta}{25-16\sin^2\theta}=20\int_{0}^{\pi/2}\frac{\cos^2(\theta)\,d\theta}{9+16\cos^2\theta} $$ and by letting $\theta=\arctan t$ the last integral boils down to: $$ 20\int_{0}^{+\infty}\frac{dt}{(1+t^2)(25+9t^2)}=\frac{\pi}{4}$$ by partial fraction decomposition.