1. ## simple pole

Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:

$\displaystyle f(z) = \frac{cos(z)}{1+z^2} + 4z$

2. Originally Posted by Dgphru
Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:

$\displaystyle f(z) = \frac{cos(z)}{1+z^2} + 4z$

Poles occur where the denominator is $\displaystyle 0 + 0i$.

So if $\displaystyle 1 + z^2 = 0$

$\displaystyle z^2 = -1$

$\displaystyle z = \pm i$.

So there are simple poles at $\displaystyle z = i$ and $\displaystyle z = -i$.

3. 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".