Unusual Application of Cauchy's Integral Formula

Question

So, I've got the following function: $f(x) = sin(\frac{1}{z-2}) + \frac{1}{z^4+1}$

I'm integrating over a circle centered at the origin with radius between 1 and 2, so some singularities are inside the circle and some are outside. I can't think of how to do this integral except with the cauchy formula, but I can't get it into the correct form.

Thanks for the help.

Answer 1

Singularities are at $z=2,\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$.

Then use Cauchy Residue Theorem, $\int_C f(z)dz=2\pi i \sum \text{Res}(z_i)$.

These residues are simple poles so I suggest using $\text{Res}(z_i)=\lim_{z\to z_i}(z-z_i)f(z)$.

Edit: The singularity at $z=2$ will never be inside your disk so do not use the residue of this singularity in your sum. The other singularities are all on the unit circle so they will always be in your disk.

Answer 2

Since the radius of the circle is between $1$ and $2$, and since the only singularity of $\sin\left(\frac1{z-2}\right)$ is outside that circle, the integral of $\sin\left(\frac1{z-2}\right)$ is equal to $0$.

So, if $r$ is the radius of the circle,\begin{align*}\int_{|z|=r}f(z)\,\mathrm dz&=\int_{|z|=r}\frac{\mathrm dz}{z^4+1}\\&=2\pi i\left(\operatorname{res}\left(\exp\left(\frac{\pi i}4\right),\frac1{z^4+1}\right)+\operatorname{res}\left(\exp\left(\frac{3\pi i}4\right),\frac1{z^4+1}\right)+\right.\\&\phantom=+\left.\operatorname{res}\left(\exp\left(\frac{5\pi i}4\right),\frac1{z^4+1}\right)+\operatorname{res}\left(\exp\left(\frac{7\pi i}4\right),\frac1{z^4+1}\right)\right)\\&=2\pi i\left(\frac1{4\exp\left(\frac{\pi i}4\right)^3}+\frac1{4\exp\left(\frac{3\pi i}4\right)^3}+\frac1{4\exp\left(\frac{5\pi i}4\right)^3}+\frac1{4\exp\left(\frac{7\pi i}4\right)^3}+\right)\\&=\frac{\pi i}2\left(\exp\left(\frac{\pi i}4\right)+\exp\left(\frac{3\pi i}4\right)+\exp\left(\frac{5\pi i}4\right)+\exp\left(\frac{7\pi i}4\right)\right)\\&=0,\end{align*}since the sum between parenthesis is the sum of the four roots of the polynomial $z^4-1$.