Limit of a complex valued integral The question is:
Compute $$\lim_{p\rightarrow0^{+}}\int_{C_p}\frac{e^{3iz}}{z^{2}-1}dz$$
Where $$C_p: z = 1 + pe^{i\theta}$$
My initial thought was to use residues, yet the poles are -1 and 1, so they're on the real line (thus the Residue Theorem does not apply). My next thought was to find some way to make the integral work with the Cauchy Integral Formula, but I can't find a way to do that since a partial fraction decomposition won't work in this case. So, I am stuck. 

Your contour $C_p$ is a circle of radius $p \to 0$ centred at the point $z=1$.  This means that there is a singularity in the contour (not on its path).  Because of this, we may use residue theorem (at singularity $z=1$) to evaluate this integral.  If 
$$f(z)=\frac{\exp(3iz)}{z^2-1}=\frac{\exp(3iz)}{(z+1)(z-1)}$$
we see that
$$\operatorname{Res}_{z=1}f(z)=\lim_{z \to 1} (z-1) \frac{\exp(3iz)}{(z+1)(z-1)} = \frac{\exp(3i)}{2}$$
Then, by residue theorem
$$\oint_{C_p} f(z)\, dz=2\pi i \operatorname{Res}_{z=1} (f(z)) = 2 \pi i \frac{\exp(3i)}{2}=\pi i \exp(3 i)$$
