# Lars Ahlfors - imaginary part of Schwarzian derivative as derivative of curvature

For my research topic I have come across an article by Lars Ahlfors, in which he computes the derivative of the curvature of a complex plane curve to be the imaginary part of the Schwarzian derivative. Quote from the article:

Let $z=f(t)$ represent a curve in the complex plane. The direction of its tangent is given by $\theta = arg(f'(t)) = Im(log(f'(t))$. The curvature measures the rate of change of $\theta$ relative to arc length and is thus $K = |f'|^{-1} Im\frac{f''}{f'}$. One more differentation shows that $\frac{dK}{ds} = |f'|^{-2}[(Im\frac{f''}{f'})' - Re\frac{f''}{f'}Im\frac{f''}{f'}] = |f'|^{-2} Im(S(f(t)).$

Here, $S(f(t))$ represents the Schwarzian derivative of $f$.

I understand the basic idea up to $K = |f'|^{-1} Im\frac{f''}{f'}$. The curvature is the derivative of $\theta$, which is evaluated as the imaginary part of the complex logarithm, and then we are using the logarithmic derivative to obtain an expression for the curvature, which we have to scale by the velocity of the curve.

However, I remain confused about the entire last line of the calculation (everything after $\frac{dK}{ds} =...$). I do not understand which rules of computation for the derivative were used to obtain the result in the first equality. Also, for the second equality, given the definition of the Schwarzian derivative as $S(f) = (\frac{f''}{f'})' - \frac{1}{2}(\frac{f''}{f'})^2$, I do not understand how $Re\frac{f''}{f'}Im\frac{f''}{f'} = Im\frac{1}{2}(\frac{f''}{f'})^2$.

The last thing I am unclear about is why derivation and taking of the imaginary part can be interchanged, i.e. why is $(Im\frac{f''}{f'})' = Im((\frac{f''}{f'})')$.

Your help is much appreciated, thank you in advance!

By the usual rule of differentiation we have $$\frac{dK}{ds}=\frac{dt}{ds}\frac{dK}{dt}=\frac{1}{|f^\prime|}\frac{dK}{dt}.$$ We want to compute $dK/dt$. By the usual rule of differentiation we have \begin{align} \frac{dK}{dt}&=\frac{d}{dt}\left(|f^\prime|^{-1}\operatorname{Im}\frac{f^{\prime\prime}}{f^\prime} \right) =\frac{\left(\operatorname{Im}\frac{f^{\prime\prime}}{f^\prime}\right)^\prime|f^\prime|-\left(\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\right)|f^\prime|^\prime}{|f^\prime|^2}\\ &=\frac{1}{|f^\prime|} \left\{\left(\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\right)^\prime -\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\cdot \frac{|f^\prime|^\prime}{|f^\prime|}\right\} . \end{align} Since \begin{align} \frac{f^{\prime\prime}}{f^\prime}&=(\log f^\prime)^\prime=(\log |f^\prime|+i\arg f^\prime)^\prime\\ &=\frac{|f^\prime|^\prime}{|f^\prime|}+i(\arg f^\prime)^\prime, \end{align} we have $$\frac{|f^\prime|^\prime}{|f^\prime|}=\operatorname{Re}\frac{f^{\prime\prime}}{f^\prime}$$ and hence $$\frac{dK}{dt}=\frac{1}{|f^\prime|} \left\{\left(\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\right)^\prime -\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\cdot \operatorname{Re}\frac{f^{\prime\prime}}{f^\prime}\right\},$$ which implies $$\frac{dK}{ds}=\frac{1}{|f^\prime|^2} \left\{\left(\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\right)^\prime -\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\cdot \operatorname{Re}\frac{f^{\prime\prime}}{f^\prime}\right\}.$$
Write $$\frac{f^{\prime\prime}}{f^\prime}=u+iv.$$ Then it is easy to check \begin{align} \frac{1}{2}\operatorname{Im}\left(\frac{f^{\prime\prime}}{f^\prime}\right)^2&=uv=\operatorname{Im} \frac{f^{\prime\prime}}{f^\prime}\cdot \operatorname{Re}\frac{f^{\prime\prime}}{f^\prime},\\ \operatorname{Im}\left(\frac{f^{\prime\prime}}{f^\prime}\right)^\prime&=\operatorname{Im}(u^\prime+iv^\prime)=v^\prime=\left(\operatorname{Im}\frac{f^{\prime\prime}}{f^\prime}\right)^\prime. \end{align} Thus we have $$\frac{dK}{ds}=\frac{1}{|f^\prime|^2}\operatorname{Im}S(f(t)).$$