$\int \frac{\exp (z)(\sin(3z)}{(z^2-2)(z^2)} dz$ on $|z|=1$

Question

So I need to calculate \begin{equation*} \int \frac{\exp(z) \sin(3z)}{(z^2-2)z^2} \, dz~\text{on}~|z|=1. \end{equation*} So I have found the singularities and residues and observed that the singularities $2^{1/2}$, $-2^{1/2}$ (the other singularity is $0$) lie outside of the radius we are looking at. My thoughts were to split in to partial fractions and the integral of the fractions corresponding with these singularities would be zero by Cauchy-Goursat theorem. But I'm unsure how to proceed with the partial fractions as my attempts keep failing. Is this even the right approach?

Thanks!

Answer 1

Consider the Taylor expansion of $\sin(3z)$

$$\sin(3z)=3z+a_1z^3+\cdots$$

Divide it by $z^2$

$$\frac{\sin(3z)}{z^2}=3/z+a_1z+\cdots$$

So, you see, we have a residue at $z=0$

Insert $z=0$ to the rest of the function, and multiply with the $3$ we get from the Taylor expansion, we will have $-\frac32$

Using the residue theorem, we get our answer: $$-3i\pi$$

Answer 2

There is a pole of order one at $z=0$. So, by the residue theorem, that residue is given by

$$\lim_{z \to 0}z\frac{e^{z}\sin(3z)}{(z^2-2)z^2}=-\frac{3}{2}$$

and the integral is $2\pi i \times (-3/2)=-i3\pi$