What is a Laurent Series and how do I find it?

Question

I'm rather new to this branch of mathematics but have been trying to learn some basics of it on my own. I know it's likely not the easiest place to start, but I've been working on calculating contour integrals. I know that the contour integral is defined as : $$ \oint f(z)\ dz = 2\pi i\sum \text{Res}\left(f,a_k\right) $$ Which would be immensely useful to me if I knew how to calculate a residue. After some research, I realized that it is the coefficient of the $-1^{\text{st}}$ term of the Laurent Series of the function of the pole for which you are trying to find the residue. I now know all of the necessary information for basic contour integration EXCEPT how to find the Laurent series of a function.

Is it like a Taylor series? Is there a simple formula for it like for a Taylor series? How does the disc/annulus of the area of $z$ affect the series? How do you know whether or not to use a disc or an annulus?

Thanks in advance, I know this may be a handful to teach in just one post, sorry for my ignorance. :/ An example would be really helpful.

Answer 1

Laurent series are for meromorphic functions, that is, functions which are analytic except they may have poles. Just as a Taylor series, you compute a Laurent series around a center. You can compute the $k$th coefficient of the Laurent series with the identity

$a_k = \oint f(z)(z-a)^{-k-1} dz$

If the function has no poles then the Laurent series coincides with the Taylor series, which implies that

$\oint \dfrac{f(z)}{(z-a)^2} dz = f'(a)$

So you can compute derivatives by integrating.

Any function analytic in a disk will have a Taylor series which converges to it in that disk. On the other hand if the function is known to be analytic in an annulus, it will have a Laurent series which converges to it in that annulus, but it need not have a convergent Taylor series, because it may have poles inside the annulus.

Usually it's easier to compute Laurent series by manipulating known Taylor series instead of using the above formula. For example the Laurent series for $\tan z$ about $z=\pi/2$ can be computed by

$\tan z= \frac{\sin z}{\cos z}= -\frac{\cos(z-\pi/2)}{\sin(z-\pi/2)} = -\frac{1+(z-\pi/2)^2/2+...}{(z-\pi/2) + (z-\pi/2)^3/3!+\dotsb}=-\frac{1}{z-\pi/2}(1-1/3(z-\pi/2)+\dotsb)=-\frac{1}{z-\pi/2}+1/3+\dotsb$

So tangent function has a single pole of order 1 at $\pi/2$.