Area theorem (conformal mapping)

Suppose that ${\displaystyle f}$ is analytic and injective in the punctured open unit disk ${\displaystyle \mathbb {D} \setminus \{0\}}$ and has the power series representation

${\displaystyle f(z)={\frac {1}{z}}+\sum _{n=0}^{\infty }a_{n}z^{n},\qquad z\in \mathbb {D} \setminus \{0\},}$

then the coefficients ${\displaystyle a_{n}}$ satisfy

${\displaystyle \sum _{n=0}^{\infty }n|a_{n}|^{2}\leq 1.}$

The idea of the proof is to look at the area uncovered by the image of ${\displaystyle f}$. Define for ${\displaystyle r\in (0,1)}$

${\displaystyle \gamma _{r}(\theta ):=f(r\,e^{-i\theta }),\qquad \theta \in [0,2\pi ].}$

Then ${\displaystyle \gamma _{r}}$ is a simple closed curve in the plane. Let ${\displaystyle D_{r}}$ denote the unique bounded connected component of ${\displaystyle \mathbb {C} \setminus \gamma [0,2\pi ]}$. The existence and uniqueness of ${\displaystyle D_{r}}$ follows from Jordan's curve theorem.

If ${\displaystyle D}$ is a domain in the plane whose boundary is a smooth simple closed curve ${\displaystyle \gamma }$, then

${\displaystyle \mathrm {area} (D)=\int _{\gamma }x\,dy=-\int _{\gamma }y\,dx\,,}$

provided that ${\displaystyle \gamma }$ is positively oriented around ${\displaystyle D}$. This follows easily, for example, from Green's theorem. As we will soon see, ${\displaystyle \gamma _{r}}$ is positively oriented around ${\displaystyle D_{r}}$ (and that is the reason for the minus sign in the definition of ${\displaystyle \gamma _{r}}$). After applying the chain rule and the formula for ${\displaystyle \gamma _{r}}$, the above expressions for the area give

${\displaystyle \mathrm {area} (D_{r})=\int _{0}^{2\pi }\Re {\bigl (}f(re^{-i\theta }){\bigr )}\,\Im {\bigl (}-i\,r\,e^{-i\theta }\,f'(re^{-i\theta }){\bigr )}\,d\theta =-\int _{0}^{2\pi }\Im {\bigl (}f(re^{-i\theta }){\bigr )}\,\Re {\bigl (}-i\,r\,e^{-i\theta }\,f'(re^{-i\theta }){\bigr )}d\theta .}$

Therefore, the area of ${\displaystyle D_{r}}$ also equals to the average of the two expressions on the right hand side. After simplification, this yields

${\displaystyle \mathrm {area} (D_{r})=-{\frac {1}{2}}\,\Re \int _{0}^{2\pi }f(r\,e^{-i\theta })\,{\overline {r\,e^{-i\theta }\,f'(r\,e^{-i\theta })}}\,d\theta ,}$

where ${\displaystyle {\overline {z}}}$ denotes complex conjugation. We set ${\displaystyle a_{-1}=1}$ and use the power series expansion for ${\displaystyle f}$, to get

${\displaystyle \mathrm {area} (D_{r})=-{\frac {1}{2}}\,\Re \int _{0}^{2\pi }\sum _{n=-1}^{\infty }\sum _{m=-1}^{\infty }m\,r^{n+m}\,a_{n}\,{\overline {a_{m}}}\,e^{i\,(m-n)\,\theta }\,d\theta \,.}$

(Since ${\displaystyle \int _{0}^{2\pi }\sum _{n=-1}^{\infty }\sum _{m=-1}^{\infty }m\,r^{n+m}\,|a_{n}|\,|a_{m}|\,d\theta <\infty \,,}$ the rearrangement of the terms is justified.) Now note that ${\displaystyle \int _{0}^{2\pi }e^{i\,(m-n)\,\theta }\,d\theta }$ is ${\displaystyle 2\pi }$ if ${\displaystyle n=m}$ and is zero otherwise. Therefore, we get

${\displaystyle \mathrm {area} (D_{r})=-\pi \sum _{n=-1}^{\infty }n\,r^{2n}\,|a_{n}|^{2}.}$

The area of ${\displaystyle D_{r}}$ is clearly positive. Therefore, the right hand side is positive. Since ${\displaystyle a_{-1}=1}$, by letting ${\displaystyle r\to 1}$, the theorem now follows.

It only remains to justify the claim that ${\displaystyle \gamma _{r}}$ is positively oriented around ${\displaystyle D_{r}}$. Let ${\displaystyle r'}$ satisfy ${\displaystyle r<r'<1}$, and set ${\displaystyle z_{0}=f(r')}$, say. For very small ${\displaystyle s>0}$, we may write the expression for the winding number of ${\displaystyle \gamma _{s}}$ around ${\displaystyle z_{0}}$, and verify that it is equal to ${\displaystyle 1}$. Since, ${\displaystyle \gamma _{t}}$ does not pass through ${\displaystyle z_{0}}$ when ${\displaystyle t\neq r'}$ (as ${\displaystyle f}$ is injective), the invariance of the winding number under homotopy in the complement of ${\displaystyle z_{0}}$ implies that the winding number of ${\displaystyle \gamma _{r}}$ around ${\displaystyle z_{0}}$ is also ${\displaystyle 1}$. This implies that ${\displaystyle z_{0}\in D_{r}}$ and that ${\displaystyle \gamma _{r}}$ is positively oriented around ${\displaystyle D_{r}}$, as required.

The inequalities satisfied by power series coefficients of conformal mappings were of considerable interest to mathematicians prior to the solution of the Bieberbach conjecture. The area theorem is a central tool in this context. Moreover, the area theorem is often used in order to prove the Koebe 1/4 theorem, which is very useful in the study of the geometry of conformal mappings.

• Rudin, Walter (1987), Real and complex analysis (3rd ed.), New York: McGraw-Hill Book Co., ISBN 978-0-07-054234-1, MR 0924157, OCLC 13093736