I'm reading the book "Teichmüller Theory and Applications to Geometry, Topology and Dynamics" by John Hubbard and I have the following question.
The proposition 7.4.4 says:
Let $\varphi,\psi$ be bounded analytic functions on $\mathbb{D}$, and for $n\in\mathbb{Z}$ set $$f(t):=\int_{\mathbb{D}}\overline{arg}(z^n+t\varphi(z))\psi(z)|dz|^2.$$
If $n\leq 1,$ the function $f$ is differentiable at $0$.
If $n=2$, the function $f$ has an asymptotic development of the form $$f(t)=f(0)+Ct\ln\frac{1}{t}+o\left(t\ln\frac{1}{t}\right),$$
with $C\neq0$ if $\varphi(0)\neq0$ and $\psi(0)\neq0$ (the function $\overline{arg}$ is defined by $\overline{arg}(z)=\frac{\overline{z}}{|z|}$).
In the proof, the part 1 it's ok.
For the part 2 we have the following equality: $$\int_{\mathbb{D}}(\overline{arg}(z^2+t\varphi(z))-\overline{arg}z^2)\psi(z)|dz|^2=\int_{\mathbb{D}}(\overline{arg}(z^2+t\varphi(0))-\overline{arg}z^2)\psi(0)|dz|^2$$ $$+\int_{\mathbb{D}}(\overline{arg}(z^2+t\varphi(z))-\overline{arg}(z^2+t\varphi(0))\psi(z)|dz|^2$$ $$+\int_{\mathbb{D}}(\overline{arg}(z^2+t\varphi(0))-\overline{arg}z^2)(\psi(z)-\psi(0))|dz|^2.$$
The author says that the last two integrals are $O(t)$. My question is about the first integral.
Suppose that $\varphi(0)\neq0$ (otherwise the integral vanishes) and make the change of variables $z=\varphi(0)^{-1/2}w$. Then $z\in \mathbb{D}$ corresponds to $|w|\in D_R$ with $R=\sqrt{1/\varphi(0)},$ and after change of variables the integral becomes
$$\int_{\mathbb{D}}(\overline{arg}(z^2+t\varphi(0))-\overline{arg}z^2)\psi(0)|dz|^2=\psi(0)\varphi(0)\int_{D_R}(\overline{arg}(w^2+t)-\overline{arg}w^2)|dw|^2.$$
This is an elliptic integral, but we can calculate its contribution to $t\ln 1/t$ in elementary terms. The integrand is bounded by $2$, so $$\int_{D_{\sqrt{|2t|}}}(\overline{arg}(w^2+t)-\overline{arg}w^2)|dw|^2\leq 4\pi t.$$
Thus it is enough to consider the integral on $\{\sqrt{|2t|}\leq |w|\leq R\}.$
So, if we set $w:=re^{i\theta}$, the author says that, with a bit of Euclidean geometry we have the inequalities
$\left|Re(\overline{arg}(w^2+t)-\overline{arg}w^2)-\frac{t\sin^22\theta}{r^2}\right|\leq 4\frac{t^2}{r^4},$ and $\int_{D_R-D_{\sqrt{|2t|}}} \frac{t^2}{r^4}\leq\frac{\pi}{2}t$.
For the first inequality, using that $$|\overline{arg}(z)-\overline{arg}(w)|\leq \left|\frac{2(z-w)}{z}\right|,$$ I only get the bound $1+\sin^2\theta\cos^2\theta$, but I have no idea how to conclude that this expression is lower than $4t^2/r^4$.
For the second inequality, computing the integral we have that $$\int_{D_R-D_{\sqrt{|2t|}}} \frac{t^2}{r^4}\leq\int_{D_R-D_{\sqrt{|2t|}}} \frac{t^2}{4t^2}=\frac{1}{4}\pi\left(\frac{1}{\varphi(0)}-2t\right).$$
What am I missing?
Sorry for the long question. Any help would be appreciated !