# Prove the existence of an analytic function

Let $\Delta=\{z\in \mathbb C: |z| < 1\}$. Let $f: \Delta \rightarrow \mathbb C$ be a one-to-one analytic function fixing the origin. Prove that there is a one-to-one analytic function $g: \Delta \rightarrow \mathbb C$ such that $[g(z)]^2=f(z^2)$. Further, show that such a function is odd.

The only idea which comes to my mind is the following (not sure whether it is the right direction though). Let $h: \Delta \rightarrow \Delta \rightarrow \mathbb C,\ z\mapsto z^2\mapsto f(z^2)$. Note that $h'(0)=0,\ h''(z)=2f'(z^2)+4z^3f''(z^2)$; since $f$ is one-to-one and analytic, its derivative never vanishes, so $h''(0)\ne 0$. So $h$ is an analytic function of $\Delta$ with a zero of order two at the origin. Then there is an analytic function $g$ defined in an open disc about the origin with a simple zero at the origin such that $[g(z)]^2=f(z^2)$.

This is the only thing I came up with, but it seems to be a little bit irrelevant since I don't see any way of extending this $g$ to the whole $\Delta$, and also the injectivity and oddness of such a $g$ are unclear.

Consider the function $f_1(z) = \frac{f(z^2)}{z^2}$. It is holomorphic on $\Delta$ and has no zeroes, so it will have a square root $h(z)$. Since $f_1$ is even, $h$ will be odd or even. Indeed, $h^2$ even implies $(h(z)-h(-z))(h(z)+ h(-z) )\equiv 0$. Since $h$ has no zeroes, it will be even. Define then $g(z) = z h(z)$. $g$ will be an odd function. You can check now that $g(z)^2 = f(z^2)$. Assume now $g(z_1) = g(z_2)$. Then $f(z_1^2) = f(z_2^2)$, and so $z_1^2 = z_2^2$. If $z_1 \ne z_2$, this implies $z_1 = - z_2$, and so $g(z_1) = - g(z_2)$. So both of them are $0$, $f(z_1^2) = 0$, so $z_1= 0 = z_2$. In conclusion, $g$ is injective and odd ( there are two solutions for $g$, clearly).

• Why "HINT"? I think this is a complete (and very elegant) answer. (By the way, I think you have a typo: $f(z_1)^2$ should read $f(z_1^2)$.)
– Cary
Oct 11 '17 at 23:54
• Just one question: as far as I understand, the existence of a square root of a non-vanishing holomorphic function is implied by the Implicit Function Theorem, which guarantees that the root exists in some neighborhood of a given point. So I can extract roots locally throughout the disc, but is there an easy way to show that the "glued root" is well-defined? (If possible, without appealing to the Monodromy Theorem.)
– Cary
Oct 12 '17 at 0:22
• @Cary: Yes, the existence of the square root, locally is no problem ( Implicit function theorem, or use some power series, or some determination ot the radical). The problem is : the local solutions should match. I can't find a way without monodromy. In fact, as you may know, if you always have a square root of a non-zero holomorphic, that guarantees the domain is iso the the unit disk. Oct 12 '17 at 0:39
• @Cary: If you don't want to use monodromy, you can mimic it. If you can define it on any rectangle with sides parallel to axes, contained in the disk, it's done. On a rectangle, divide it into small pieces, over which you can define a root. Start piecing together ( think - reverse process of eating a bar of chocolate, piece by piece), That is in fact "baby monodromy". Oct 12 '17 at 0:44
• $\frac{f_1'}{f_1}$ is holomorphic and hence so is $\log f_1$ and $f_1^{1/2}$ @Cary Oct 12 '17 at 1:53

You want $g$ to be a branch of $\sqrt{f(z^2)}$. You have shown this to be possible in a neighbourhood of $0$. It is certainly possible in a neighbourhood of any other point of $\Delta$ (since $f$ is one-to-one and $f(0)=0$, $f(z) \ne 0$ there). The Monodromy Theorem should show that you get analytic continuation to $\Delta$. Moreover there are exactly two possible $g$ (since there are two choices for the square root in a neighbourhood of some $z \ne 0$).

Now if $g(z)$ is one solution, so are $-g(-z)$ and $g(-z)$. Thus one of these must be $g(z)$, i.e. $g$ is either even or odd. But by looking at the order of the zero at the origin, it can't be even.