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is a counterclockwise rotation of \(f\) about \(z=0\) through \(\alpha\) radians. Expressing it as power series, we get that if
\[f(z)=z+\sum_{n=2}^{\infty}a_{n}z^{n}\quad\mbox{then}\quad g(z)=z+\sum_{n=2}^{ \infty}b_{n}z^{n},\]
where \(b_{n}=a_{n}e^{i(n-1)\alpha}\). Therefore
\[|b_{n}|=|a_{n}||e^{i(n-1)\alpha}|=|a_{n}|.\]
These functions
\[k_{\lambda}(z)=\frac{z}{(1-\lambda z)^{2}}=\sum_{n=1}^{\infty}n\lambda^{n-1}z ^{n}\quad(\lambda\ \ \mbox{a constant and}\ \ |\lambda|=1)\]
are, in fact, the only functions for which \(|a_{n}|=n\) for some (and hence for all) \(n\), up to rotations. In the following example \(f(z)\) is a polynomial of degree \(n\) and we try to find a certain bound on \(|a_{n}|\).
Figure 3.45: Koebe function
**Example 117**.: Given
\[f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots+a_{n}z^{n}\]
a polynomial of degree \(n\), if \(f\) is univalent in \(D(0;1)\), then \(|a_{n}|\leq\dfrac{1}{n}\). To see this, consider
\[f^{\prime}(z)=1+2a_{2}z+\cdots+na_{n}z^{n-1}=na_{n}(\dfrac{1}{na_{n}}+\cdots+z ^{n-1}).\]
Now use the Fundamental Theorem of Algebra to factor the above polynomial of degree \(n-1\) as
\[f^{\prime}(z)=na_{n}(z-c_{1})(z-c_{2})\cdots(z-c_{n-1})\]
where \(c_{1},c_{2},\ldots,c_{n-1}\) are complex roots. However, \(f\) is univalent on \(D(0;1)\), so \(f^{\prime}(z)\) has no roots in \(D(0;1)\) and thus \(|c_{k}|>1\) for all \(k\). Since \(f^{\prime}(0)=1\),
\[1=|f^{\prime}(0)|=|na_{n}||c_{1}||c_{2}|\cdots|c_{n-1}|\geq|na_{n}|\]
as claimed.
The most important property of univalent functions is the famous Riemann mapping theorem which we state below again. See also Theorem 127.
**Theorem 149** (Riemann Mapping Theorem).: _Let \(G\) be a simply connected region (“without holes”) with \(G\neq\mathbb{C}\). Then there exists a one-to-one conformal mapping \(f:G\to D(0,1)\) with \(f^{-1}:D(0,1)\to G\) is also conformal._
The Riemann mapping theorem is an existence result; it does tell you how you can find such a mapping. It is not clear at all that it is possible to construct a conformal map from a region with complicated boundary onto a nice region such as \(D(0;1)\) or vice versa. The strength of the Riemann mapping theorem is the reason why the Bieberbach conjecture stayed a main problem in geometric function theory till de Branges’ proof.
### Bieberbach’s Area Theorem
The Bieberbach conjecture came to be postulated after the proof of the following area theorem. The basic way to obtain Bieberbach like inequalities is to relate the power series coefficients to the area of some region in the plane. Let \(\mathcal{U}\) denote the class of functions \(g\) which are holomorphic and univalent in \(\Sigma=\{z:|z|>1\}\) and have Laurent expansion
\[g(z)=z+b_{0}+\frac{b_{1}}{z}+\cdots=z+\sum_{n=0}^{\infty}\frac{b_{n}}{z^{n}}.\]
Proof
: Let \(R=\mathbb{C}\setminus g(\Sigma)\) be the complement of the image set \(g(\Sigma)\). Our goal is to calculate the area of \(R\) in terms of the \(b_{n}\). More precisely, we set \(R(r)=\mathbb{C}\setminus\{g(z):|z|>r\}\) and find the area bounded by the simple closed curve \(\gamma(r)\) (here by \(\gamma(r)\) we denote the boundary of \(R(r)\).) Then we define (Figure 3.46):
\[\operatorname{area}R=\lim_{r\to 1^{+}}\operatorname{area}R(r).\]
Let \(g(z)=u+iv=w\); clearly,
\[\operatorname{area}R(r)=\iint_{R(r)}du\,dv=\frac{1}{2i}\iint_{R(r)}d\overline {w}\,dw,\]
since \(du\,dv=\frac{1}{2i}d\overline{w}\,dw\). Now, we use Green’s theorem to write
\[\operatorname{area}R(r)=\frac{1}{2i}\int_{\gamma(r)}\overline{w}\,dw=\frac{1 }{2i}\int_{|z|=r}\overline{g(z)}g^{\prime}(z)\,dz.\]
Next we parametrize \(z=re^{it}\), \(dz=ire^{it}dt\) to express the last integral as:
\[\operatorname{area}R(r)=\frac{1}{2}\int_{0}^{2\pi}re^{it}g^{\prime}r(e^{it}) \overline{g(re^{it})}dt.\]
Using the power series for \(g\) and \(g^{\prime}\),
\[\frac{1}{\pi}\operatorname{area}R(r)=\frac{1}{2\pi}\int_{0}^{2\pi}\left(re^{ it}+\sum_{n=1}^{\infty}nb_{n}r^{-n}e^{-int}\right)\left(re^{-it}+\sum_{n=0}^{ \infty}\overline{b_{n}}r^{-n}e^{int}\right)dt.\]
Figure 3.46: Area of R
The last expression can be simplified using the orthogonality of distinct powers of \(e^{it}\) to
\[\frac{1}{\pi}\text{area}\,R(r)=r^{2}-\sum_{n=1}^{\infty}r^{-2n}\,n\,|b_{n}|^{2}.\]
Noting that \(\text{area}\,R(r)\geq 0\), and thus the partial sums \(\sum_{n=1}^{m}r^{-2n}\,n\,|b_{n}|^{2}\leq r^{2}\) for every \(m>0\) and letting \(r\to 1^{+}\) yield the claim that \(\sum_{n=1}^{m}n|b_{n}|^{2}\leq 1,\text{ for }m=1,2,\dots\).
This theorem gives us an immediate corollary, useful in proving the Bieberbach theorem:
**Corollary 27**.: _If \(g(z)\) is in \(\mathcal{U}\), then \(|b_{1}|\leq 1\), with equality if and only if \(g\) has the form_
\[g(z)=z+b_{0}+\frac{b_{1}}{z}\qquad|b_{1}|=1.\]
### The Koebe \(\frac{1}{4}\)-Theorem
Another well-known result, coming from the case \(n=2\) in the Bieberbach conjecture is known as the Koebe \(\frac{1}{4}\) -theorem. This theorem states that a univalent function \(f\) on \(D(0;1)\) with the conditions \(f(0)=0\) and \(f^{\prime}(0)=1\) is one such that
\[D(0,\frac{1}{4})\subseteq f\left(D(0;1)\right).\]
Equality occurs for those class of functions that give equality in the Bieberbach conjecture. Namely, equality can hold if and only if \(f\) is a rotation of the Koebe function. In other words, the range of each function \(f\in S\) contains some disc centered at the origin. Already in 1907, Koebe conjectured that \(r\leq\frac{1}{4}\), with the maximum attained by the Koebe function. Later in 1916, Bieberbach proved the theorem, showing that this constant cannot be improved.
**Remark 80**.: Progress to prove the Bieberbach conjecture occurred in several directions; some can be grouped as follows:
* \(|a_{n}|\leq n\) for a particular \(n\),
* \(|a_{n}|\leq n\) for subclasses of \(S\),
* \(|a_{n}|\leq C\) for sufficiently large \(C\)
One of the results of the first type is the third coefficient theorem, \(|a_{3}|\leq 3\), due to K. Loewner. His proof was completely different than Bieberbach’s second coefficient theorem and he used a partial differential equations method which was useful to de Branges’s proof of the general conjecture. The Bieberbach conjecture also had been proved for certain subclasses of S. For example, the subclass of starlike functions, and the subclass of functions with real coefficients [20]. As an evidence of the difficulty of solving Bieberbach conjecture one can use the basic but powerful tools of analysis (Cauchy’s inequality) to show for \(f\in S\),
\[|a_{n}|\leq\frac{e^{2}}{4}n^{2}\quad n=2,3,\ldots\]
See [55] for details. Note that the above inequality does not even get the order of growth of the coefficients right. As an example for the third group of results we present the following work of Littlewood.
### Littlewood Theorem
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