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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?

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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 $\frac{t_5}{t_6}=(-2n-n^2)(z-z_0)^{-2}=(1-m^2)(z-z_0)^{-2}$.

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  • $\begingroup$ 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. $\endgroup$ – Martin R Dec 5 '19 at 8:35
  • $\begingroup$ @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. $\endgroup$ – Ewan Delanoy Dec 5 '19 at 8:40
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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.

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

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