How do you find the Laurent series of f(z) = 1/(z² + 1)?

I have a mental block and cannot remember the right procedure of getting this function into the right form.

Thanks for any help.

Long division will do the job, won't it?

Laurent? i think a Taylor series will suffice for this function, if you expand around z=0. (A Taylor series around z=0 is known as a McLaurin Series). To find it, you have to calculate the kth derivative at 0.

or um, if you really want a Laurent series, why not expand around z=-1?

then the Laurent series is 1/(z+1)²

(in other words, the function you give is already in Laurent form)

Edit: Oops, i misread your function, i thought you had written 1/(z+1)². Let me try to answer your question for 1/(z²+1)

OK, well now i guess you could use the formula for the Laurent series about a

it is

f(z) = Σ [1/(2πi)∫ f(z)/(z-a)ⁿ⁺¹ dz](z-a)ⁿ

but only suckers use that. better to write the function in terms of stuff you know the Taylor series for.

that is 1/(z²+1) = 1/2(1/(1-iz) - 1/(1+iz)) by partial fractions

then write 1/(1-iz)=i/z * 1/(1 - 1/iz) and 1/(1+iz)=1/iz * 1/(1+1/iz) since you want to expand where |z| > 1 (otherwise the Taylor/McLaurin series for this function exists and is the whole Laurent series for |z| < 1)

since |z|>1, you can write Taylor series for those, which is just the formula for the sum of a geometric series. i.e. 1/(1 - 1/iz) = &Sigma;(1/iz)ⁿ

then add them, and you're done

Thanks Lethe.

It is the first step in the partial fraction expanding that gets me!

that is 1/(z&sup2;+1) = 1/2(1/(1-iz) - 1/(1+iz)) by partial fractions

How did you do this? What were you thinking when you did this? In the exam today I tried all sorts of other tricks I know to get it into its partial fraction form but none of them worked. Obviously I need to subject myself to more exotic fractions that the ones I'm used to.

Any help explaining how you did this would be appreciated.

Thanks Lethe.

It is the first step in the partial fraction expanding that gets me!



How did you do this? What were you thinking when you did this? In the exam today I tried all sorts of other tricks I know to get it into its partial fraction form but none of them worked. Obviously I need to subject myself to more exotic fractions that the ones I'm used to.

Any help explaining how you did this would be appreciated.

partial fractions are always got the same way. you know that the factors in the denominator must make up the denominators of the partial fractions, so write something like

1/(ax+b)(cx+d)(ex+f) = A/(ax+b) + B/(cx+d) + C/(ex+f)

and we want to solve for A, B, and C. so then multiply by (ax+b)(cx+d)(ex+f):

1 = A(cx+d)(ex+f) + B(ax+b)(ex+f) + C(ax+b)(cx+d)

then expand, and collect powers of x:

1 = (Ace + Bae + Cac) x² + [A(de+cf)+B(be+af)+C(bc+ad)]x + (Aef+Bbf+Cbd)

if this equality is to hold true for all x, then the coefficients must match:

Ace + Bae + Cac = 0
A(de+cf)+B(be+af)+C(bc+ad) = 0
Aef+Bbf+Cbd = 1

three equations, three unknowns. solve for A, B, and C.

if the factor in the denominator is greater than degree one (say q(x) is quadratic), then it looks like this:

1/(ax+b)q(x) = A/(ax+b) + (Cx+D)/q(x)

in general, the numerator can have degree up to one less than the degree of the denominator.

also, if one of the factors is repeated then it goes like this:

1/(ax+b)²(cx+d) = A/(ax+b) + B/(ax+b)² + C/(cx+d)


and that's partial fractions in a nutshell. you use them to integrate ugly rational functions, and a lot of other places as well.

so in the example at hand, i wrote

1/(z²+1)=1/(z+i)(z-i) = A/(z+i) + B/(z-i)

1 = A(z-i) + B(z+i)

A+B = 0
i(B-A) = 1

so B=-i/2 and A = i/2

so 1/(z²+1)=1/2(1/(z+i)-1/(z-i))

then i divided through by iz to get something that i could use in the geometric summation formula

Yep, that did it! My mistake was ignoring the fact that x&sup2; + 1 = (z+i)(z-i) when you factorize it. Oh well some easy marks lost again. :(

Thanks for your help Lethe.