Compute the integral $\int_{\gamma}^{}\frac{2z}{(z+1)(z+3)}$ Compute the integral $$\int_{\gamma}^{}\frac{2z}{(z+1)(z+3)}dz$$, where $\gamma(t)=2e^{it}$, $0\leq t\leq2\pi$.
My initial idea was that the function has singularities at $z=-1, \; -3$ and then it´s residue would be $-1,\; 3$ respectively and entire integral would have the value of $12\pi i$
The second idea I had was to continue with parameterization:
$$ \int_{\gamma}^{}\frac{2z}{(z+1)(z+3)} dz = \int_{0}^{2\pi}\frac{2*2e^{it}}{(2e^{it}+1)(2e^{it}+3)}*2ie^{it}dt$$
but this is the part where I am stuck. I would appreciate any kind of help with this.
 A: 1st you need to find the singularities, which are at $z=-1$ and $z=-3.$ Then you need to check which of those are inside your curve. It is obvious that $-1 \in int(\gamma)$ but $-3 \notin int(\gamma).$ Consider now the function $g(z)=\frac{2z}{z+3}.$ This function is Holomorphic inside your curve and in the point $z_0=-1.$ So we will apply Cauchys integral formula for the function g at $z_0.$
$$2\pi i g(-1)=\int_{\gamma}\frac{g(z)}{z+1}dz.$$
The integrand on the right hand side is your function, because $\frac{g(z)}{z+1}=\frac{\frac{2z}{z+3}}{z+1}=\frac{2z}{(z+1)(z+3)}.$
So in order to finish you need to calculate $g(-1).$
$$2\pi i g(-1)=\int_{\gamma}\frac{g(z)}{z+1}dz$$
$$\Longrightarrow \int_{\gamma}\frac{2x}{(z+2)(z+1)}dz=2\pi i \frac{-2}{2}=-2\pi i. $$
And you are done!
A: There is no need to re-parameterize the integral. Whenever you have a rational integrand on a circular contour, this is a very good place to apply the residue theorem.
By the residue theorem, the integral is simply equal to $2\pi i$ times the sum of the function's residue at all of the poles inside the contour. This would just be the pole at $z=-1$ (since $z=-3$ is outside of the contour)
The residue is given by $$\lim_{x\to-1}(z+1)\cdot \frac{2z}{(z+1)(z+3)}=-1$$
Hence, the integral is $$2\pi i\cdot -1=\boxed{-2\pi i}$$
