I have a complex analysis course this semester and we were introduced to series 2 weeks ago. My instructor has shown us power series, Maclaurin series, Laurent series and some other types of series. She showed us some examples of series and one of them was,

Expand $f(z)=\sin(z)$ as Maclaurin series. The result for this example was,


and then we moved to the next example and it was,

Expand $f(z)=\frac{\sin(z)}{z}$ around $z=0$ as Laurent series. I was expected her to use,

$$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}, (R_1\lt|z-z_0|\lt R_2)$$

definition but she used the result we've found in the last example and divided both sides by $z$ to find the result. Like below.

$$f(z)=\frac{\sin(z)}{z}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!}=\sum _{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\frac{z^{2n+1}}{z}$$


What I wonder is, she used expansion of $\sin(z)$ as Maclaurin series to find expansion of function $\frac{\sin(z)}{z}$ around $z=0$ as Laurent series. Is this correct? If so, why?

And my second question is if we would try to find expansion of $\frac{\sin(z)}{z}$ around $z=0$ as Laurent series using the definition I wrote above, would we find the same result?


  • 1
    $\begingroup$ 1. Yes it's correct, though I'm not sure why you should believe me over your professor. 2. Yes, every meromorphic function has a unique Laurent series in a specific annulus in its domain, so the same result arises regardless of method. $\endgroup$ Jun 6, 2021 at 20:08
  • $\begingroup$ @AndrewD.Hwang sorry for misunderstanding. I couldn't understand whether the answer of my instructor correct or not by your comment. $\endgroup$ Jun 6, 2021 at 20:13
  • $\begingroup$ There is no constant term in the expansion for $sin(z)$, so dividing $by $z$ leaves it as a Maclaurin series. $\endgroup$ Jun 6, 2021 at 20:44
  • 4
    $\begingroup$ $z=0$ is a removable singularity. If we define $f(z)=\sin(z)/z$ for $z\neq 0$ and $f(z)=1$ if $z=0$, then $f$ is an entire function. The Laurent series is just the Taylor series. $f(z) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots $. $\endgroup$ Jun 6, 2021 at 22:16
  • 1
    $\begingroup$ For sums use \sum not \Sigma. $\endgroup$
    – K.defaoite
    Jun 6, 2021 at 23:03

1 Answer 1


Power (or Taylor or Maclaurin) series are special cases of Laurent series.

By their uniqueness properties (which are often underplayed), any process that produces any such series as an output produces the same thing (where we mean that the center point is the same, and inner radius of an annulus is the same, etc.)

So, yes, a holomorphic function on a disk has Laurent series which is actually a power series: there are no negative-degree terms.

(A function holomorphic on an annulus need not extend to a function with an isolated singularity at the center, and certainly need not extend to a meromorphic function on that punctured disk, to have a Laurent expansion on the annulus.)

So, yes, $\sin z = z+ z^3/3!+...$ is both the power series expansion at $0$ and the Laurent expansion at $0$... with no negative terms.

A-priori, sure, we know that ${\sin z\over z}$ is meromorphic on the complex plane punctured at $0$, so has a Laurent expansion. But/and the singularity at $0$ is removable, so it has a power series expansion (special type of Laurent). Further, we can obtain that power series (or Laurent series, in more complicated cases) by dividing the power series for $\sin z$ by $z$. Bingo. (Since the power series for $\sin z$ has no zero-degree term, the quotient of that power series by $z$ is still a power series... not a Laurent series.)

If you wanted the Laurent expansion of ${\cos z\over z}$ at $0$, divide the power series $\cos z = 1 + z^2/2!+\ldots$ by $z$ to have $$ {\cos z\over z} \;=\; {1\over z} + {z\over 2!} + \ldots $$ Here, there is a negative-index term.


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