Let $0<r<R$ .
$(a)$ Show that :
$$\frac{1}{2πi}\oint_{|z|=r} \frac{R+z}{(R-z)z} \,dz = 1$$
$(b)$ Using $(a)$, show that :
$$\frac{1}{2 \pi} \int_{-\pi}^{\pi} \frac{R^2-r^2}{R^2 + r^2 - 2rR\cosθ}dθ =1$$
My question is regarding $(a)$ (I just posted the full exercise so it can be presented fully for someone who may read it in the future). Now, this exercise is a part of a handout for revision before the semester exams. Fact is, I've never handled something similar (except some inequality-proofs regarding complex-circles and Radius-regarded expressions in integrals), so I did some theory-research to see how such a question could be handled.
I am almost sure that such a proof can be carried out by using the Residue-related theorems and especially :
$$\int_C f(z)dz = 2\pi i Res(f,z_0)$$
where $C$ any circle in the form of $|z-z_0|=r$, positively oriented with $0<r<R$.
Now, in my case, our circle is $C:|z|=r$ which means $z_0=0$.
So :
$$\oint_{|z|=r} \frac{R+z}{(R-z)z} \,dz = 2\pi i Res\bigg(\frac{R+z}{(R-z)z},0\bigg) $$
$$\Leftrightarrow $$
$$\frac{1}{2πi}\oint_{|z|=r} \frac{R+z}{(R-z)z} \,dz = Res\bigg(\frac{R+z}{(R-z)z},0\bigg)$$
Now it's obvious this is the way it's getting proved, but I would like to ask some help or guidance on understanding how the Residue expression above "works" and how can I find it. My question pretty much comes down on how to express the $Res\bigg(\frac{R+z}{(R-z)z},0\bigg)$.
Any help would be greatly appreciated !
Thanks for your time !
To find the value of the integral using the residue theorem, you first have to find the singularities of the integrand inside the contour. You're correct in looking only at $z=0$, but you should probably say something about why you don't need to consider $R-z$. To find the value of the residue at $z=0$ look at $$\lim_{z\to0}z\frac{R+z}{(R-z)z}=\lim_{z\to0}\frac{R+z}{(R-z)}=1$$