Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:
$\displaystyle f(z) = \frac{cos(z)}{1+z^2} + 4z $
I dont get it sighs.. please help me
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".