# How can I solve the curve integral of $\frac{cos(z)}{z^2}$ with the unit circle as curve? [closed]

I tried to make a partial fraction decomposition but that didn't work. Can someone show me how to do it with the residue theorem?

• What are the poles and residues? Please show what you have done so far. – Kavi Rama Murthy Feb 23 at 8:45
• $\cos(z)/z^2$ is even. – user10354138 Feb 23 at 8:48
• There is a double pole at z = 0 and I think the residue is 0. (it's the first time I try to use this theorem) – syximak Feb 23 at 8:50

By Cauchy's integral formula: $$\int_{\gamma}\frac{\cos(z)}{z^2}dz=\frac{2\pi i}{1!}\cos'(0)=0$$ where $$\gamma$$ is the unit circle curve.