simple pole

**Dgphru** · May 19th 2010, 10:42 PM

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

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

I dont get it sighs.. please help me

**Prove It** · May 19th 2010, 10:51 PM

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:

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

I dont get it sighs.. please help me

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

So if $1 + z^2 = 0$

$z^2 = -1$

$z = \pm i$ .

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

**HallsofIvy** · May 20th 2010, 03:19 AM

More generally, simple poles (poles of order 1) occur where the Laurent series for a function has a " $(z- a)^{-1}$ " 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 $\frac{cos(z)}{z^2+ 1}+ 4z= \frac{cos(z)}{(z- i)(z+ i)}+ 4z$ .

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. $\frac{1}{z+ i}$ is analytic everywhere except at z= -i, so it has a Taylor series expansion about z= i and therefore so does the product $\frac{cos(z)}{z+i}$ . Multiplying that Taylor series by $\frac{1}{z- i}$ introduces a term with $(z- i)^{-1}$ . 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".

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