Please help me solve this complex integral. $$\oint_C \frac{\cos(z-a)}{(z-a)}\mathrm{d}z$$

Such that $a\in \Bbb R^2$ and $C$ is a single closed curved defined by $|z-a|=\frac{|a|}{2}$

Here $z=x+iy$ is a complex number. Please solve the above integral. Thanks
Let $$f(z) = \cos(z - a)$$

Then you use the Cauchy Integral Formula:

$$f(z_0) = \frac1{2 \pi i}\oint_C \frac{f(z)}{(z-z_0)}dz$$

Let $z_0 = a$. Therefore this becomes:

$$f(a)=\frac1{2 \pi i}\oint_C \frac{\cos(z-a)}{(z-a)}dz = \cos(a-a) = \cos(0) = 1$$

Rearranging for your integral:

$$\oint_C \frac{\cos(z-a)}{(z-a)}dz = 2 \pi i \cos(0) = 2 \pi i$$
Let $$f(z) = \cos(z - a)$$

Then you use the Cauchy Integral Formula:

$$f(z_0) = \frac1{2 \pi i}\oint_C \frac{f(z)}{(z-z_0)}dz$$

Let $z_0 = a$. Therefore this becomes:

$$f(a)=\frac1{2 \pi i}\oint_C \frac{\cos(z-a)}{(z-a)}dz = \cos(a-a) = \cos(0) = 1$$

Rearranging for your integral:

$$\oint_C \frac{\cos(z-a)}{(z-a)}dz = 2 \pi i \cos(0) = 2 \pi i$$