I'm trying to find the solution to a definite integral using the Residue Theorem. I've currently found the poles of the complex function $f(z)$ but am unsure of how to classify those poles in order to apply the limit definition of the residue.
Of the poles, $2$ are within the unit circle in the complex plane but I'm unsure of the order of these poles. I'm evaluating the integral at the top of the image.
Best Answer
You are almost done. I usually take a slightly different $f$:$$f(z)=\frac1z\cdot\frac1{1+\left(\frac{z+1/z}2\right)^2}=\frac{4 z}{z^4+6 z^2+1}.$$The roots of $z^4+6 z^2+1$ are, as you know, $\pm i\sqrt{3\pm2\sqrt2}$, of which only $\pm i\sqrt{3-2\sqrt2}$ have absolute value smaller than $1$. Since $z^4+6 z^2+1$ has degree four, each of these roots is a simple root and so it's a simple pole of $f$. And so\begin{align}\operatorname{res}_{z=\pm i\sqrt{3-2\sqrt2}}f(z)&=\frac{\pm4i\sqrt{3-2\sqrt2}}{4\left(\pm i\sqrt{3-2\sqrt2}\right)^3+12\left(\pm i\sqrt{3-2\sqrt2}\right)}\\&=\frac1{2\sqrt2}.\end{align}So, your integral is equal to$$2\pi\left(\frac1{2\sqrt2}+\frac1{2\sqrt2}\right)=\pi\sqrt2.$$