Solving the functional equation $\tau \left(\frac{-1}{z}\right) = - \tau(z)$

Question

Let $\mathbb{H} \subset \mathbb{C}$ be the upper half plane. Find $\tau: \mathbb{H} \to \mathbb{C} $, holomorphic and non-constant, satisfying $\tau \left( \frac{-1}{z} \right) = - \tau(z)$.

There is a very good answer already. But since the question was put on hold, I'll add some context. This question came up in the context of Klein's $j$ - invariant. If you change coordinates from $z \to w=z-i$ and expand in $w$, you'll see from symmetry arguments that the first two coefficients must be $0$. From this, I'm wondering if $j$ is locally the square of a coordinate around $i$. This makes sense if you look at the fundamental domain for $\mathbb{H}$ mod the $SL(2, \mathbb{Z})$ action, and note that $j$ defines a complex analytic structure on it.

Answer 1

Take any $f$ holomorphic on the upper half-plane and let $$\tau(z)=f(z)-f(-1/z).$$

Answer 2

We have that $$\tau'(-1/z)=-z^2\tau'(z)$$

The function $\sigma(z)=1/z$ has the following property: $$\sigma(-1/z)=-z=-z^2\sigma(z)$$

So a solution could come from $\tau'(z)=1/z$. Then $\tau(z)=C+\log z$ where $C$ is a complex constant.

Then $$C+\log(-1/z)=-C-\log(z)$$ $$C=-\frac{\log(z)+\log(-1/z)}2=-i\frac{\arg(z)+\pi-\arg(z)}2=-i\pi/2$$

Thus, $$\tau(z)=-i\pi/2 + \log z=\log(-iz)$$

Answer 3

Letting $z = i\frac{1-x}{1+x}$ turns the functional equation into $g(x) = -g(-x)$, and the solutions to those are the functions of the form $g(x) = xh(x^2)$.

Going back to the original setting, the solutions are exactly the functions of the form $\tau(z) = \frac{i-z}{i+z} f((\frac{i-z}{i+z})^2)$ where $f$ is any holomorphic function on the unit disk.

Answer 4

EqWorld is our right partner.

In fact this functional equation belongs to the form of http://eqworld.ipmnet.ru/en/solutions/fe/fe1121.pdf.

The general solution is $\tau(z)=C\left(z~,-\dfrac{1}{z}\right)$ , where $C(u,v)$ is any antisymmetric function.