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.

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