# How to visualise Laurent Series Coefficients?

I'm trying to think about this same as Taylor series $$\displaystyle{f(z) = \sum_{k = 0}^{\infty} a_k\,(z-z_0)^k}$$, wit coefficients $$a_k = \dfrac{f^{k}(z_0)}{k!}$$. You can easily imagine how $$f(z)$$ is approximated at $$z_0$$ by adding more and more polynomial parts of higher dimension (First it's a line, then a parabola etc.)

I have problems to extend this idea on Laurent series, learning about this right now: $$f(z) = \displaystyle{f(z) = \sum_{k = 0}^{\infty} a_k\,(z-z_0)^k} + \sum_{k = 0}^{\infty} b_j(z-z_0)^{-j}$$ with $$a_k = \dfrac{1}{2\,\pi\,i} \int_C\dfrac{f(z)}{(z-z_0)^{k+1}}\,\mathrm{dz}$$ and $$b_j = \dfrac{1}{2\,\pi\,i} \int_C\dfrac{f(z)}{(z-z_0)^{-n+1}}\,\mathrm{dz}$$.

I do know Laurent series is used when the point $$z_0$$ where you build your series from is not analytic expressing it with the help of coefficients $$b_j$$. Also Laurent Series just seems to work in a domain $$n_1 < \vert z\vert This might make it hard to visualise the whole process. The best I can do is thinking about $$f(z)$$ as a 2 Dimensional function (magnitude of $$f(z)$$ over the imaginary-real-plane) approximated by a 2 Dimensional Taylor Series. Just that not analytic points $$z_0$$ like singularities are included. This does not explain why the series is restricted to $$n_1<\vert z \vert where it only converges.

• In your notation you should have $n_1 < |z-z_0| < n_2$ instead of $n_1 < |z| < n_2$. Jul 2, 2021 at 21:12

## 1 Answer

One way to visualise the meaning of Laurent series is to use the Fourier series.

Let us look at the values of function $$f(z)$$ along the circle given by $$|z-z_0|=r$$, where we asume that this circle lies in the region of the convergence of the given Laurent series. If $$f(z) = \sum_{k=-\infty}^\infty a_k (z-z_0)^k$$ then $$f(z_0+re^{i\phi}) = \sum_{k=-\infty}^\infty a_k r^k e^{ik\phi}$$ That means that if $$a_k$$ are the coefficients of the Laurent series of function $$f(z)$$, then $$a_kr^k$$ are the coefficients of the Fourier series (in the exponential form) of the function $$g(\phi) = f(z_0+re^{i\phi})$$. And just as Fourier series is the decomposition of a periodic function into trigonometric functions, the Laurent series is a decomposition of a holomorphic functions into function that "oscilate" around point $$z_0$$: function $$(z-z_0)^0 = 1$$ has no oscilations, function $$(z-z_0)$$ has one oscilation around the circle, function $$(z-z_0)^k$$ has $$k$$ oscilations, function $$(z-z_0)^{-k}$$ also has $$k$$ oscilations, but in a sense "in the oposite direction".

As for the the understanding of the convergence of the Laurent series, the division $$f(z) = \sum_{k=0}^\infty a_k(z-z_0)^k + \sum_{k=1}^{\infty} a_{-k} (z-z_0)^{-k}$$ Notice that if $$|z-z_0|$$ is too big, then $$|z-z_0|^{k}$$ may grow too fast for the series $$\sum_{k=0}^\infty a_k(z-z_0)^k$$ to converge; on the other hand, if $$|z-z_0|$$ is to small, then $$|z-z_0|^{-k}$$ may not go down fast enough for the series $$\sum_{k=1}^{\infty} a_{-k} (z-z_0)^{-k}$$ to converge. If both too big and too small values of $$|z-z_0|$$ are not allowed, you get a condition of the form $$n_1 < |z-z_0| < n_2$$.

• Great analogy. Of course it goes back to Fourier Series.
– Leon
Jul 3, 2021 at 12:22