Proving $\frac1{2\pi} \int_0^{2\pi} \frac{R^2-r^2}{R^2-2Rr\cos\theta+r^2} d\theta =1$ by integrating $\frac{R+z}{z(R-z)}$ without residue theorem.

We could avoid complex analysis altogether

$$I = \frac{1}{\pi}\int_0^\pi \frac{R^2-r^2}{R^2+r^2 - 2Rr\cos\theta}\:d\theta$$

$$ = \frac{1}{\pi}\int_0^\pi \frac{R^2-r^2}{R^2+r^2\left(\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}\right) - 2Rr\left(\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\right)}\:d\theta$$

$$\frac{1}{\pi}\int_0^\pi \frac{(R-r)(R+r)}{(R-r)^2\cos^2\frac{\theta}{2}+(R+r)^2\sin^2\frac{\theta}{2}}\:d\theta = \frac{2}{\pi}\int_0^\pi \frac{\left(\frac{R-r}{R+r}\right)\cdot\frac{1}{2}\sec^2\theta\:d\theta}{\left(\frac{R-r}{R+r}\right)^2+\tan^2\frac{\theta}{2}}$$

$$= \frac{2}{\pi}\tan^{-1}\left[\left(\frac{R-r}{R+r}\right)\tan\frac{\theta}{2}\right]\Biggr|_0^{\pi^-} = 1$$

with the assumption that $|R|\neq|r|$