I'd like to evaluate following integral with contour integration $$\int^{\infty }_{-\infty}\dfrac {z^3\sin az}{z^4+4}dz$$ and I think the best way to solve is to recognize it is equal to the imaginary part,
$$\Im\int^{\infty }_{-\infty}\dfrac {z^3e^{iaz}}{z^4+4}dz$$
I would like to use the residue theorem to calculate the above contour integral, noting the poles at $ \dfrac{i\pi}{4}$ and $ \dfrac{3i\pi}{4}$.
I've learned a theorem saying that for $f(z)=\dfrac{p(z)}{q(z)}$ where $p(z_0) \neq 0$, $q(z_0)= 0$, $q'(z_0) \neq 0$, then the residue of f(z) at $z_{0}$ is $\dfrac{p(z_0)}{q'(z_{0})}=\dfrac{e^{iaz}}{4}$ for this problem.
Since $z=\exp\left(\dfrac{i\pi}{4}\right)$ and $z=\exp\left(\dfrac{3i\pi}{4}\right)$ I do not know how to calculate the residues.
For the $\dfrac{\pi}{4}$ part, I get that the residue at that pole is $\dfrac{\exp\left({ia\exp\left({\dfrac{i\pi}{4}}\right)}\right)}{4}$. Is that right?
How can I calculate that value? And the analogous value at the other pole?
You have the poles wrong. Solving $z^4 + 4 = 0$ gives you $z = \pm 1 \pm i$. Note that the choice of contour depends on the sign of $a$. Rremember to use Jordan's lemma to justify your computation.
All the poles are simple, so computing the relevant (depending on the choice of contours, thus on the sign of $a$) residues can be done as: $$ \operatorname{Res}\limits_{z = \pm 1 \pm i}\bigg( \frac{z^3 e^{iaz}}{z^4+4} \bigg) = \frac{z^3 e^{iaz}}{4z^3} \bigg|_{z=\pm 1 \pm i} = \frac14 e^{ia(\pm 1 \pm i)} $$