Forming a Laurent Series for 4cos(z*pi) / (z-2i)

[Dissonance in E]

How does one form a laurent series about the point z0 = 2i
for the function:

4cos(z*pi) / (z-z0).

Could one take advantage of the power series
1 / (z - z0)
1 / (z - 2i)

SUMMATION q^n
= SUMMATION (2i)^n
= 1 + 2i -4 -8i . . . . .

and somehow integrate the rest of the original function into the above results?

 

[HallsofIvy]

Well, first you have to handle that "cos(z)"!
I suspect it will be better to use cos(z)= cos(z- 2i+ 2i)= cos(z- 2i)cos(2i)- sin(z- 2i)sin(2i). Expand those as power series, then divide by z- 2i. That will give just a single $(z- 2i)^{-1}$ and all the rest will be non-negative powers.