Compute $\oint_{\partial D_3(0)}\frac{\cos(z+4)}{z^2+1}dz$

Question

I want to compute $$\oint_{\partial D_3(0)}\frac{\cos(z+4)}{z^2+1}dz$$ without using the residue theorem. My plan was to do a partial fractians decomposition: $$\frac{\cos(z+4)}{z^2+1}=\frac{i}{2}\cos(z+4)\left(\frac{-1}{z-i}+\frac{1}{z+i}\right)$$And then I want to use the Cauchy integral formula: $$\oint_{\partial D_3(0)}\frac{\cos(z+4)}{z^2+1}dz=\frac{i}{2}\oint\frac{-\cos(z+4)}{z-i}dz+\frac{i}{2}\oint\frac{\cos(z+4)}{z+i}dz$$ $$=-\frac{i}{2}\cos(4+i)+\frac{i}{2}\cos(4-i)$$

Is this correct? If not, how do I approach it?

Answer 1

Yes, it is fine, except that you forgot to multiply by $2\pi i$ at the last line. The answer is $\pi\bigl(\cos(4+i)-\cos(4-i)\bigr)$.

Answer 2

You can avoid partial fraction decomposition by noting that

$$\oint_{\partial D_3(0)}\frac{\cos(z+4)}{z^2+1}dz = \oint_{\partial \color{blue}{D_1(i)}}\frac{\frac{\cos(z+4)}{z+i}}{z-i}dz + \oint_{\partial \color{blue}{D_1(-i)}}\frac{\frac{\cos(z+4)}{z-i}}{z+i}dz$$ $$= 2\pi i \left( \frac{\cos(i+4)}{i+i} + \frac{\cos(-i+4)}{-i-i}\right) = \pi(\cos(4+i)- \cos(4-i)) $$