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

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.

• It's a slightly lengthy and complex (!) subject to be thoroughly treated in this site, I think. Grab a book in complex analysis, any decent book will do, or wai until you reach that level in your class. In short, yes: it's kindda Taylor series but with negative exponents. Oct 13, 2016 at 19:18
• +1 for @DonAntonio's comment. "Read a book" might seem like cheap advice, but it's really the best possible in this context. Take the time to properly learn the fundamentals of a subject. If you try to jump ahead to the advanced machinery without a solid foundation, you'll never really understand what you're doing. You're already dwelling under a minor misconception: the formula for a contour integral that you wrote is not how they are defined. Rather, it's a very important important theorem that includes assumptions on the function $f$ (en.wikipedia.org/wiki/Residue_theorem). Oct 13, 2016 at 19:38
• If $f(z)= \sum_{n=-\infty}^\infty a_n z^n$ converges absolutely for $0 < |z| < 2$ then $\oint_{|z|= 1} f(z)dz= \ ?$ Oct 13, 2016 at 20:53

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 $\pi/2$.

• Thanks so much for this answer! It helped me with so many of my questions(even those that weren't in the original post). You have been very helpful, and I plan to do as @Glare has said and find a good book to read. Thanks again! Oct 13, 2016 at 23:43