Does there exist a univalent map \[ f:\mathbb C\setminus\Delta\to\mathbb C \] of the form \[ f(z)=z+\sum_{n=1}^{\infty}\frac{b_n}{z^n} \] with \[ b_2=\frac1{\sqrt2}? \]
Solution
No such univalent map exists.
1. The area theorem would force every coefficient except \(b_2\) to vanish.
For a normalized exterior univalent map \[ f(z)=z+\sum_{n\ge1}\frac{b_n}{z^n}, \] the area theorem gives \[ \sum_{n\ge1} n|b_n|^2\le1. \] If \[ b_2=\frac1{\sqrt2}, \] then the \(n=2\) term alone contributes \[ 2|b_2|^2=2\cdot\frac12=1. \] Therefore every remaining nonnegative term must vanish: \[ b_n=0\qquad(n\ne2). \] Hence the only possible map is \[ f(z)=z+\frac{1}{\sqrt2\,z^2}. \]
2. This forced candidate is not univalent on the exterior disk.
Its derivative is \[ f'(z)=1-\frac{\sqrt2}{z^3}. \] Thus \(f'(z)=0\) whenever \[ z^3=\sqrt2. \] Every such critical point has modulus \[ |z|=2^{1/6}>1, \] so it lies in the exterior domain. A holomorphic injective map has nonzero derivative everywhere in its domain. Hence this candidate is not univalent, contradicting the assumption.
Therefore no normalized exterior univalent map can have \(b_2=1/\sqrt2\).