## Area under holomorphic maps

Let $f$ be a map from $(x,y)\in \mathbb{R}^2$ to $(a,b)\in \mathbb{R}^2$. The area form $dA=dx\wedge dy$ gives the Jacobian $dA=da\wedge db= J(x,y)dx\wedge dy$, where $J(x,y)=a_xb_y- a_yb_x$.

Now consider the complex setting, where $\displaystyle dA=\frac{i}{2} dz\wedge d\bar z$. Let $f$ be a map from $z\in \mathbb{C}$ to $w\in \mathbb{C}$. Then $\displaystyle dA=\frac{i}{2} dw\wedge d\bar w= \frac{i}{2}f'(z)\overline{f'(z)} dz\wedge d\bar z$. So this time the Jacobian $J(z)$ becomes $f'(z)\overline{f'(z)}$.

Suppose $\displaystyle f(z)=\sum_{n\ge 0} a_n z^n$ is a holomorphic map on the unit disk $D$. Then
$\displaystyle J(z)=\sum_{n,m\ge 0}nm a_n \bar a_m z^{n-1} \bar z^{m-1}$, the area of $f(D)$ is $\displaystyle \int_D J_f(z) dA$.

Using polar coordinate, we have $dA= rdr\, d\theta$, $\displaystyle z^{n-1} \bar z^{m-1}=r^{n+m-2}e^{i\theta(n-m)}$,
and $\displaystyle \int_D r^{n+m-2}e^{i\theta(n-m)} rdr\, d\theta= 0$ if $n\neq m$, and $=\frac{\pi}{n}$ if $n=m$.

So $\displaystyle |f(D)|=\sum_{n\ge 0} n^2 |a_n|^2\cdot \frac{\pi}{n}=\pi \sum_{n\ge 0} n |a_n|^2$.

Let $p(x)=a_d x^d+\cdots + a_0$ be an irreducible polynomial with integer coefficients, $\alpha$ be an irrational number such that $f(\alpha)=0$. Then $\alpha\in D(d,\epsilon)$ for some $\epsilon>0$.

Proof. Let $M=\sup\{|f'(x)|: |x-\alpha|\le 1\}$. For each rational number $p/q$, we have $f(p/q)\neq 0$. Eliminating the denominator, we have $a_d p^d+\cdots a_0 q^d \neq 0$. Each item being integral, we see that $|a_d p^d+\cdots a_0 q^d|\ge 1$ and hence $|f(p/q)|\ge q^{-d}$. On the other hand, $|f(p/q)|=|f(p/q)-f(\alpha)|\le M |\alpha-p/q|$. Combining them, we have $|\alpha-p/q|\ge \epsilon q^{-d}$, where $\epsilon=1/M$.