Find $\oint_\Gamma \frac{\cos z}{z(z^2+8)}dz$ where $\Gamma$ is a positevly oriented unit circle.

Question

Find $$\oint_\Gamma \frac{\cos z}{z(z^2+8)}dz$$ where $\Gamma$ is a positevly oriented unit circle.

My attempt:

$$\oint_\Gamma \frac{\cos z}{z(z^2+8)}dz=\oint_\Gamma \frac{\cos z}{z(z-\sqrt8 i)(z+\sqrt8 i)}$$

Now I apply residue theorem to get:

$$2\pi i (Res(f,0)+Res(f,\sqrt8 i)+Res(f,-\sqrt8 i))=2\pi i(1/8-\frac{\cos(\sqrt8 i)}{8})$$

But I don't like this answer. Is is possible to compute this integral using Cauchy Integral Formula?

Answer 1

Note that $\pm\sqrt{8}i$ are not in the unit disk. Hence the answer is simply $$2\pi i Res(f,0)=\frac{i\pi}{4}.$$