Let $h:V\to V$ be a holomorphic automorphism fixing $0$ of $V$ open in $\mathbb{C}$. Let $\bar{D_r}\subset V$
$h(\partial \bar{D_r})$ is a convex curve iff $\operatorname{arg}(z)+\operatorname{arg}(h'(z))+\pi/2$ is an increasing function of $\operatorname{arg}(z)$
The derivative of this function with respect to $\operatorname{arg}(z)$ is given by $\operatorname{Re}(\frac{z h''}{h'})+1$.
What does $\operatorname{arg}(z)+\operatorname{arg}(h'(z))+\pi/2$ mean geometrically? How can I differentiate this function with respect to $\operatorname{arg}(z)$?
How should I prove above statement? Any hints or references are appreciated. Thank you
Let $$\partial D_r:\quad t\mapsto z(t)=re^{it}\qquad(0\leq t\leq2\pi)$$ be the standard parametric representation of $\partial D_r$. Then $$\gamma:\quad t\mapsto p(t):=h\bigl(re^{it}\bigr)$$ is a parametric representation of the image curve $\gamma:=h(\partial D_r)$. One computes $$p'(t)=h'(re^{it})ir e^{it},\quad p''(t)=-h''(re^{it})r^2 e^{2it}-h'(re^{it})re^{it}\ .$$ Now for the geometry. The tangent direction of the image curve at the point $p(t)$ is given by $$\theta(t)={\rm arg}\bigl(p'(t)\bigr)={\rm arg}\bigl(h'(z)\bigr)+{\rm arg}(z)+{\pi\over2}\>,\qquad z:=re^{it}\ .\tag{1}$$ Here we have used that ${\rm arg}$ converts products into sums. If we want $\gamma$ to be strictly convex then we want the function $t\mapsto\theta(t)$ strictly increasing with $t$, hence $\theta'(t)>0$. Write the first part of $(1)$ in the form $$\theta(t)={\rm arg}\bigl(p'(t)\bigr)={\rm Im}\bigl(\log p'(t)\bigr)\ .$$ Then we obtain $$\theta'(t)={\rm Im}\left(p''(t)\over p'(t)\right)={\rm Im}\left(-{h''(re^{it})r e^{it}\over ih'(re^{it})}-{1\over i}\right)={\rm Re}\left({zh''(z)\over h'(z)}+1\right)\ .$$