Give an example of a sequence of degree-three rational maps \[ f_n:\widehat{\mathbb C}\to\widehat{\mathbb C} \] such that \[ f_n(S^1)\subset S^1 \] and \(f_n(z)\to z\) uniformly on \(S^1\). For your example, compute \[ \int_{S^1}\frac{f_n'(z)}{f_n(z)}\,dz. \]
Solution
For \(c\in(0,1)\), set \[ B_c(z)=\frac{z-c}{1-cz}. \] Fix \[ a=\frac12, \qquad b_n=\frac12+\frac1{4n}, \] and define \[ \boxed{f_n(z)=z\frac{B_a(z)}{B_{b_n}(z)}}. \]
1. Each \(f_n\) is a rational map of degree \(3\).
Writing the function as one quotient gives \[ f_n(z) =\frac{z(z-a)(1-b_nz)}{(1-az)(z-b_n)}. \] The numerator has degree \(3\) and the denominator degree \(2\). Their zeros are respectively \[ 0,\ a,\ 1/b_n \qquad\text{and}\qquad b_n,\ 1/a. \] Since \(a\neq b_n\) and \(b_n\neq a^{-1}\), there is no common factor. Thus the induced rational map on \(\widehat{\mathbb C}\) has degree \(3\).
2. One has \(f_n(S^1)\subset S^1\).
For \(|z|=1\) and real \(0<c<1\), \[ |z-c|=|1-cz|, \] so \(|B_c(z)|=1\). Hence on \(S^1\), \[ |f_n(z)| =|z|\frac{|B_a(z)|}{|B_{b_n}(z)|} =1. \]
3. The maps \(f_n\) converge uniformly to the identity on \(S^1\).
Since \(b_n\to a\) and the denominators \(1-b_nz\) stay uniformly away from zero on \(S^1\), the functions \(B_{b_n}\) converge uniformly to \(B_a\) on \(S^1\). Moreover \(|B_{b_n}|=1\) there, so \[ \frac{B_a}{B_{b_n}}\longrightarrow1 \] uniformly on \(S^1\). Therefore \[ f_n(z)=z\frac{B_a(z)}{B_{b_n}(z)}\longrightarrow z \] uniformly on \(S^1\).
4. Inside the unit disk, \(f_n\) has two zeros and one pole, all simple.
From the factored expression in 1, the zeros in \(\Delta\) are \[ z=0,\qquad z=a, \] while the only pole in \(\Delta\) is \[ z=b_n. \] The remaining zero \(1/b_n\) and pole \(1/a\) lie outside the closed unit disk.
5. Consequently\[ \boxed{ \int_{S^1}\frac{f_n'(z)}{f_n(z)}\,dz=2\pi i.} \]
By the argument principle for a meromorphic function, \[ \frac1{2\pi i} \int_{S^1}\frac{f_n'(z)}{f_n(z)}\,dz =N-P, \] where \(N\) and \(P\) are the numbers of zeros and poles in the disk, counted with multiplicity. By 4, \(N=2\) and \(P=1\). Hence the integral is \(2\pi i\).