Complex analysis - calculate integral

Question

Calculate integral:

$$\oint\limits_{|z|=2018}\dfrac{\pi^{\frac{1}{z}}}{(1+3z^2)^{2018}}\,\mathrm dz $$

Can anyone help me with calculating this integral? I don't have any idea how to do it. On lectures was another examples and i have a problem with it.

Answer 1

All the singularities of $f(z)=\dfrac{\pi^{\dfrac{1}{z}}}{(1+3z^2)^{2018}}$ ($z=0$ and $z=\pm\frac{1}{\sqrt 3}i$) are contained by the path $|z|=2018$. So it suffices to find the residue of $\dfrac{1}{z^2}f(\dfrac{1}{z})$ in $z=0$. $$\dfrac{1}{z^2}f(\dfrac{1}{z})=z^{4034}\dfrac{\pi^z}{(z^2+3)^{2018}}$$which has an obvious residue of zero in $z=0$. Therefore the above integral is 0 either.

Answer 2

Use the residue theorem. Your function has $3$ singularities: at $0$ and at $\pm\frac i{\sqrt{3}}$. Compute the residue at each one of them.