# Laurent expansion of the Schwarzian derivative

The following question is from Ahlfors's Complex Analysis, in a section about Laurent series.

Question. The expression $$\{f,z\}=f'''(z)/f'(z)-(3/2)(f''(z)/f'(z))^2$$ is called the Schwarzian derivative of $$f$$. If $$f$$ has a multiple zero or pole, find the leading term in the Laurent development of $$\{f,z\}$$. [Answer: If $$f(z) = a(z - z_0)^m + · · ·$$, then $$\{f,z\} = (1/2)(1 - m^2)(z - z_0)^{-2} + · · ·$$. ]

I first tried to solve this by direct computation, but it has gone very messy, so I think that there should be another method. Is there a better way to do this?

Note that you may run into problems with your formula if you do not make more assumptions on $$f$$ : for example, the Schwarzian derivative is undefined when $$f$$ is constant.
Also, if $$m=0$$, $$a$$ does not appear in $$f'$$ or the Schwarzian derivative, so the formula is false. Below, I show that the formula holds when $$m\neq 0$$ or $$1$$.

Let $$g=f'$$. Then $$\lbrace f \rbrace=\frac{g''}{g}-\frac{3}{2}\bigg(\frac{g'}{g}\bigg)^2=\frac{N}{D}$$ where $$N=2gg"-3(g')^2$$ and $$D=2g^2$$. Then , we have successively :

The principal term in the Laurent expansion of $$f$$ is $$t_{-1}=a(z-z_0)^m$$.

The principal term in the Laurent expansion of $$g$$ is $$t_0=t_1'=b(z-z_0)^{n}$$ where $$b=am$$ and $$n=m-1$$.

The principal term in the Laurent expansion of $$g'$$ is $$t_1=t_0'=bn(z-z_0)^{n-1}$$.

The principal term in the Laurent expansion of $$g''$$ is $$t_2=t_1'=bn(n-1)(z-z_0)^{n-2}$$ (unless $$m=2$$).

The principal term in the Laurent expansion of $$2gg''$$ is $$t_3=2t_0t_2=2b^2n(n-1)(z-z_0)^{2n-2}$$ (unless $$m=2$$).

The principal term in the Laurent expansion of $$3(g')^2$$ is $$t_4=t_1^2=3b^2n^2(z-z_0)^{2n-2}$$.

The principal term in the Laurent expansion of $$N$$ is $$t_5=t_3-t_4=b^2(-2n-n^2)(z-z_0)^{2n-2}$$ (note that this stays true even when $$m=2$$, because in this case one of the summands is zero, but the total formula stays the same).

The principal term in the Laurent expansion of $$D$$ is $$t_6=2t_0^2=2b^2(z-z_0)^{2n}$$.

The principal term in the Laurent expansion of $$\lbrace f \rbrace$$ is $$\frac{t_5}{t_6}=(-2n-n^2)(z-z_0)^{-2}=(1-m^2)(z-z_0)^{-2}$$.

• As I understand it, $f$ is assumed to have a multiple zero or pole at $z=z_0$, therefore it cannot be constant, and $m$ cannot be $0$ or $1$ (or $-1$). The formula does hold for $m=2$ though, I don't see why you excluded that value. – Martin R Dec 5 '19 at 8:35
• @MartinR Corrected, thanks. Note that there are a few adjustements to make for the proof to work on the $m=2$ case, see my update. – Ewan Delanoy Dec 5 '19 at 8:40

It becomes a bit easier if you write the Schwarzian derivative in the equivalent form $$\{f,z\} = \left( \frac{f''(z)}{f'(z)}\right)' - \frac 12 \left( \frac{f''(z)}{f'(z)}\right)^2 \, .$$ At a multiple zero or pole of $$f$$ we have $$f(z) = a (z-z_0)^m + \ldots$$ for $$z \to z_0$$ with $$a \ne 0$$ and an integer $$m \ne -1, 0, 1$$. Then \begin{align} f'(z) &= am (z-z_0)^{m-1} + \ldots \\ \implies \frac{f''(z)}{f'(z)} &= \frac{m-1}{z - z_0} + O(1) \end{align} so that \begin{align} \left( \frac{f''(z)}{f'(z)}\right)' &= -\frac{m-1}{(z - z_0)^2} + \ldots \\ \left( \frac{f''(z)}{f'(z)}\right)^2 &= \frac{(m-1)^2}{(z - z_0)^2} + \ldots \end{align} and finally \begin{align} \{f,z\} &= \left(-(m-1) - \frac12 (m-1)^2 \right) \frac{1}{(z - z_0)^2} + \ldots \\ &= \frac{\frac 12(1-m^2)}{(z - z_0)^2} + \ldots \end{align} For a multiple zero or pole is $$1-m^2 \ne 0$$, so that this gives the leading term of the Schwarzian derivative for $$z \to z_0$$.

Remark: Since the Schwarzian derivative does not change if a constant is added to $$f$$, one can generalize the statement slightly:

If $$f$$ is meromorphic in a neighbourhood of $$z_0$$ and $$f(z_0) =a$$ (which can be finite or $$\infty$$) with multiplicity $$m \ge 2$$ then $$\{f,z\} = \frac{\frac 12(1-m^2)}{(z - z_0)^2} + \ldots$$
for $$z \to z_0$$.

It is also easy to see that $$\{f,z\}$$ is holomorphic at all points where $$f$$ takes a value (finite or infinite) with multiplicity one.

• +1, this is definitely an improvement over my answer – Ewan Delanoy Dec 5 '19 at 8:49