simple pole

Sep 2009
41
0
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
 

Prove It

MHF Helper
Aug 2008
12,896
5,001
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
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

MHF Helper
Apr 2005
20,249
7,909
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".