If is a univalent (1-1 analytic) function with domain the unit disc such that \(f(z) = z + \sum_{n=2}^{\infty} a_n z^n\), then prove that
\[ g(z) = \sqrt{f(z^2)} \]
is an odd analytic univalent function on the unit disc.
The source statement reads “If is a univalent (1-1 analytic) function”, omitting the name of the function; the formula \(f(z) = z + \sum_{n=2}^{\infty} a_n z^n\) and the definition of \(g\) show that the function meant is \(f\).
Solution
Since \[ f(w) = w+\sum_{n=2}^{\infty}a_nw^n, \] define \[ \phi(z) \coloneqq 1+\sum_{n=2}^{\infty}a_n z^{2n-2}. \]
1. The function \(\phi\) is analytic, even, and nonvanishing on the unitdisk, and\[ f(z^2)=z^2\phi(z). \]
Substituting \(w=z^2\) in the power series for \(f\) gives \[ f(z^2) = z^2+\sum_{n=2}^{\infty}a_nz^{2n} = z^2\phi(z). \] The displayed power series for \(\phi\) contains only even powers of \(z\), so \(\phi\) is analytic and even, with \(\phi(0)=1\).
If \(z\neq0\) and \(\phi(z)=0\), then \[ f(z^2)=0=f(0). \] Since \(f\) is univalent, this would imply \(z^2=0\), a contradiction. At \(z=0\), one has \(\phi(0)=1\). Hence \(\phi\) has no zeros in the unit disk.
2. There is an analytic function \(h\) on the unit disk such that\[ h(z)^2=\phi(z) \]and \(h(0)=1\).
The unit disk is simply connected, and step 1 shows that \(\phi\) is a nonvanishing analytic function there. Hence \(\phi\) admits an analytic square root. Choosing the sign of that square root so that its value at \(0\) is \(1\) gives \(h\).
3. The function \(h\) is even.
Because \(h^2=\phi\) and \(\phi\) is even, \[ h(-z)^2 = \phi(-z) = \phi(z) = h(z)^2. \] The function \(h\) is nonvanishing, so \[ q(z)\coloneqq\frac{h(-z)}{h(z)} \] is analytic and satisfies \(q(z)^2=1\). Thus its values lie in the discrete set \(\{1,-1\}\). Since the unit disk is connected, \(q\) is constant. At \(z=0\), \[ q(0)=1, \] so \(q\equiv1\) and therefore \(h(-z)=h(z)\).
4. The analytic square-root branch\[ g(z)\coloneqq zh(z) \]satisfies\[ g(z)^2=f(z^2) \]and is odd and analytic on the unit disk.
Steps 1 and 2 give \[ g(z)^2 = z^2h(z)^2 = z^2\phi(z) = f(z^2). \] Thus \(g\) is an analytic choice of the square root in the statement. Since \(h\) is analytic, so is \(g\). By step 3, \[ g(-z) = -z\,h(-z) = -z\,h(z) = -g(z), \] so \(g\) is odd.
5. The function \(g\) is univalent on the unit disk.
Suppose \[ g(z_1)=g(z_2). \] Squaring and using step 4 gives \[ f(z_1^2)=f(z_2^2). \] Since \(f\) is univalent, \[ z_1^2=z_2^2, \] so either \(z_1=z_2\) or \(z_1=-z_2\).
In the second case, oddness from step 4 and the assumed equality give \[ g(z_1) = g(-z_1) = -g(z_1), \] so \(g(z_1)=0\). Then \[ 0=g(z_1)^2=f(z_1^2)=f(0). \] Univalence of \(f\) implies \(z_1^2=0\), hence \(z_1=z_2=0\). Thus in every case \(z_1=z_2\).
6. Q.E.D.
Step 4 proves that the intended square-root branch is analytic and odd, and step 5 proves that it is univalent.