How to integrate $\frac{\sin(z)}{(z^2 + 1/2)^2}$ over the unit circle? How can I integrate $\frac{\sin(z)}{(z^2 + 1/2)^2}$?
$(z^2 + 1/2)^2$ isn't in a form where I can use the Cauchy Integral Formula, and we haven't covered using residues yet in my class so I'm assuming that's not the way to go (it's also fairly ugly to try to do so in this case).
 A: You can apply Cauchy's integral theorem by expanding the factor $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ in partial fractions. You can obtain the partial fraction expansion without having to solve any equations as follows. The singularities are at $z = \alpha^{\pm} = \pm \frac{i}{2}\sqrt{2}$. The partial fraction expansion of the function $\frac{1}{z^2 + \frac{1}{2}}$ is thus given by:
$$\frac{1}{z^2 + \frac{1}{2}} = \sum_{\pm}\left(\lim_{z\to\alpha^{\pm}}\frac{z-\alpha^{\pm}}{z^2+\frac{1}{2}}\right)\frac{1}{z-\alpha^{\pm}} = \frac{i}{2}\sqrt{2}\left(\frac{1}{z+\frac{i}{2}\sqrt{2}} - \frac{1}{z-\frac{i}{2}\sqrt{2}}\right) $$
Squaring both sides gives:
$$\frac{1}{\left(z^2 + \frac{1}{2}\right)^2} = -\frac{1}{2}\left(\frac{1}{\left(z+\frac{i}{2}\sqrt{2}\right)^2} + \frac{1}{\left(z-\frac{i}{2}\sqrt{2}\right)^2}\right)+\frac{1}{\left(z+\frac{i}{2}\sqrt{2}\right)\left(z-\frac{i}{2}\sqrt{2}\right)}$$
Around the pole at $z = -\frac{i}{2}\sqrt{2}$ we have the expansion:
$$\frac{1}{\left(z^2 + \frac{1}{2}\right)^2} =-\frac{1}{2}\frac{1}{\left(z+\frac{i}{2}\sqrt{2}\right)^2} +\frac{i}{2}\sqrt{2}\frac{1}{z+\frac{i}{2}\sqrt{2}}+\text{nonsingular terms}$$
Around the pole at $z = \frac{i}{2}\sqrt{2}$ we have the expansion:
$$\frac{1}{\left(z^2 + \frac{1}{2}\right)^2} =-\frac{1}{2}\frac{1}{\left(z-\frac{i}{2}\sqrt{2}\right)^2} -\frac{i}{2}\sqrt{2}\frac{1}{z-\frac{i}{2}\sqrt{2}}+\text{nonsingular terms}$$
If we subtract from $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ the four terms from the two expansions then this will cancel out all the singularities (we're left with only removable singularities), therefore this must yield a polynomial. But since this tends to zero at infinity, it must be zero. The function $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ is thus equal to the sum of the four singular terms in the two expansions.
