when dealing with Laurent series I am confused about how to compute negative coefficients.

Laurents theorem states (in my course at least) that if f is differentiable on $D(z_0, R)/ \{z_0\}$ R>0. then $\exists a_n\in \Bbb C s.t.f(\zeta)=\sum_{n=-\infty}^{\infty}a_n(\zeta-z_0)^n$, $\zeta \in D(z_0,R)/\{z_0\}$

where $a_n=1/2\pi i\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz$

combining this with cauchy's integral formula for derivatives gives $a_n=\frac{f^n(z_0)}{n!}$.

I tried to compute the Laurent series of $\frac{1}{z^2}$ but I dont understand how to do this for $a_{-1}$ , as $a_{-1}=f^{-1}(0)/-1!$ and I'm not sure how to compute this.

Does $f^{-1}(0)$ simply mean the inverse here ?

Does $-1!=-(1!)$, i.e. does $-n!=-(n!).$

  • $\begingroup$ You can use the Residue formula $\endgroup$ – Quoka Apr 22 '18 at 14:46
  • $\begingroup$ @MathUser_NotPrime We don't learn the residue formula until next year, I just want to know how to compute Laurent series which is the topic we're covering at the minute. $\endgroup$ – excalibirr Apr 22 '18 at 14:47
  • $\begingroup$ You are trying to find the Laurent series about what point? i.e. what is $z_0$ $\endgroup$ – Quoka Apr 22 '18 at 14:49
  • $\begingroup$ @MathUser_NotPrime I'm trying to find it at the singular point $z_0=0$ to to decide whether or not it's an essential singularity. $\endgroup$ – excalibirr Apr 22 '18 at 14:52

As mentioned in another answer, the Laurent series of $z^{-2}$ is simply $z^{-2}$. We now show it using your formula. First, note that $$ a_n = \frac{1}{2\pi i}\int_{C(0,1)}\frac{z^{-2}}{z^{n+1}}dz = \frac{1}{2\pi i}\int_{C(0,1)}z^{-3-n}dz $$ So if $-3-n \geq 0$, the $a_n = 0$ (by Cauchy's theorem). Thus, $a_n = 0$ for all $n\leq -3$. Now, if $n=-2$ then we may compute $$ a_{-2} = \frac{1}{2\pi i}\int_{C(0,1)}z^{-1}dz = \frac{1}{2\pi i}\int_0^{2\pi}\left(e^{i\theta}\right)^{-1}ie^{i\theta}d\theta =1 $$ Now, if $n \geq -1$ then \begin{align*} a_{n} = \frac{1}{2\pi i}\int_{C(0,1)}z^{-(n+3)}dz &= \frac{1}{2\pi i}\int_0^{2\pi}\left(e^{i\theta}\right)^{-(n+3)}ie^{i\theta}d\theta\\ &= \frac{1}{2\pi}\int_0^{2\pi}e^{-(n+2)i\theta} d\theta\\ &= \left.\frac{1}{2\pi}\frac{e^{-(n+2)i\theta}}{-(n+2)i}\right|_0^{2\pi} = 0 \end{align*}

  • $\begingroup$ this is perfect , I know how to compute them now :) $\endgroup$ – excalibirr Apr 22 '18 at 15:29
  • $\begingroup$ a small additional question the fact that $a_n=0$ for $\forall n \neq-2 \Rightarrow$ implies this is not an essential singularity and because the order of f at $z_0 < 0$ means its not removable. we do have though that that the order is negative means that we have a pole of order -2 at $z_0$ correct ? $\endgroup$ – excalibirr Apr 22 '18 at 15:37
  • $\begingroup$ At $z_0$ (which in this case is $0$) we have a pole of order 2, not $-2$. But yes, it is not an essential singularity nor is it removable $\endgroup$ – Quoka Apr 22 '18 at 17:05

The formula of $$ a_n=\frac{1}{2\pi i}\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz = \frac{f^n(z_0)}{n!}. $$ works only for $n > 0$. Note that $f^n(z)$ is the $n$-th derivative of $f$.

For $n \le 0$, it depends on the singularity of $f$ at $z_0$.

For example, if $f(x) = \frac{g(x)}{(x-x_0)^k}$ where $g(x)$ is holomorphic, then $$ a_n=\frac{1}{2\pi i}\int_{C(z_0,r)}\frac{g(x)}{(z-z_0)^{n+k+1}}dz. $$ Thus for $n > -(k+1)$, one can apply Cauchy's integral formula. For $n \le -(k+1)$, the integrand becomes analytic, which makes $a_n = 0$.


Cauchy's integral formula is for derivatives, that is, for things like $f^{(n)}(a)$, with $n\in\mathbb{Z}_+$. It doesn't make sense to try to apply it with negative $n$.

And the Laurent series of $\frac1{z^2}$ centered at $0$ is simply $z^{-2}$ or, if you prefer, $\sum_{n=-\infty}^{+\infty}a_nz^n$ with $a_{-2}=0$ and $a_n=0$ if $n\neq-2$.

  • $\begingroup$ How does one calculate the Laurent series step by step ? Like we can do for Taylor expansions . $\endgroup$ – excalibirr Apr 22 '18 at 15:03
  • $\begingroup$ @exodius That's too broad. Do you have a specific example in mind? $\endgroup$ – José Carlos Santos Apr 22 '18 at 15:05
  • $\begingroup$ say for example $\frac{1}{z^2+1}$ $\endgroup$ – excalibirr Apr 22 '18 at 15:07
  • $\begingroup$ @exodius You use the fact that$$\frac1{z^2+1}=\frac12\left(\frac1{1-\frac zi}+\frac1{1+\frac zi}\right),$$together with the fact that$$|z|<1\implies\frac1{1-z}=1+z+z^2+z^3+\cdots$$and that$$|z|>1\implies\frac1{1-z}=-z^{-1}-z^{-2}-\cdots$$ $\endgroup$ – José Carlos Santos Apr 22 '18 at 15:15
  • $\begingroup$ Ah I think I understand now, thank you for your detailed answer :) $\endgroup$ – excalibirr Apr 22 '18 at 15:30

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