More generally, simple poles (poles of order 1) occur where the Laurent series for a function has a "\(\displaystyle (z- a)^{-1}\)" term but no other negative exponents. (The order of a pole is the "highest" negative exponent in the Laurent series expansion.)

Here, the function is \(\displaystyle \frac{cos(z)}{z^2+ 1}+ 4z= \frac{cos(z)}{(z- i)(z+ i)}+ 4z\).

cos(z) is an analytic function for all z so its "Laurent" series is actually its Taylor series about that point and has no negative exponents. \(\displaystyle \frac{1}{z+ i}\) is analytic everywhere except at z= -i, so it has a Taylor series expansion about z= i and therefore so does the product \(\displaystyle \frac{cos(z)}{z+i}\). Multiplying that Taylor series by \(\displaystyle \frac{1}{z- i}\) introduces a term with \(\displaystyle (z- i)^{-1}\). Finally, just adding "4z" to the series does not change that. This function has a pole of order 1 at z= i.

For z= -i, just swap "i" and "-i".