# simple pole

#### Dgphru

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

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

#### Prove It

MHF Helper
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 $$\displaystyle z = -i$$.

• HallsofIvy and Dgphru

#### HallsofIvy

MHF Helper
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".