$f\left(\frac{2z}{1+z^2}\right)=\left(1+z^2\right)f(z)$, solve $f$.

Problem

Given $f$ is smooth and defined on $(-1,0)\cup (0,1)$, $\lim_{z\to 0}f(z)$ exists and it satisfies

$$f\left(\frac{2z}{1+z^2}\right)=\left(1+z^2\right)f(z).\tag {*}$$

Solve $f$.

First attempt

I tried to extend $f$ to all real number arguments by substituting $z=\frac{1}{t}$,

$$f\left(\frac{2t}{1+t^2}\right)=\left(1+\frac1{t^2}\right)f\left(\frac1t\right).\tag {1.1}$$

Therefore, $\forall z\in(-\infty,0)\cup (0,\infty)$,

$$f(z)=\frac1{z^2}f\left(\frac1z\right).\tag {1.2}$$

Let $g(z)=zf(z)$,

$$g(z)=g\left(\frac1z\right).\tag {1.3}$$

How can this be continued?

Second attempt

I notice $$\tanh2x=\frac{2\tanh x}{1+\tanh^2x}.\tag {2.1}$$

Let $g(z)=zf(z)$ again, then

$$g\left(\frac{2z}{1+z^2}\right)=2g(z).\tag {2.2}$$

Put $z=\tanh x$,

$$g(\tanh2x)=2g(\tanh x).\tag {2.3}$$

Write $h(x)=g(\tanh x)$,

$$h(2x)=2h(x).\tag {2.4}$$

$$\therefore h(x)=cx, \forall c\in \mathbb R.$$

$\therefore g(x)$ is the inverse of $\tanh\dfrac xc$.

$$\therefore f(z)=\frac{c \text{ arctanh } z}{z}, \forall c\in \mathbb R.\tag {**}$$

As putting $\tanh x$ is not the general method, I want to see other ways of solving the problem.

Third attempt

$$f\left(\frac{2z}{1+z^2}\right)=\left(1+z^2\right)f(z).$$

Let $h(z)=zf(z)$, $$h\left(\frac{2z}{1+z^2}\right)=2h(z).\tag {3.1}$$

Assume the power series

$$h(z)=c_0+\sum_{k=1}^\infty c_k z^k.\tag {3.2}$$

I try to find the coefficients of $h$.

Then $$h\left(\frac{2z}{1+z^2}\right)=c_0+\sum_{k=1}^\infty c_k \left(\frac{2z}{1+z^2}\right)^k.\tag {3.3}$$

$$h\left(\frac{2z}{1+z^2}\right)=c_0+\sum_{k=1}^\infty c_k 2^k z^k \sum_{n=0}^\infty \binom{n+k-1}{n} \left(-z^2\right)^n$$ $$h\left(\frac{2z}{1+z^2}\right)=c_0+\sum_{k=1}^\infty \sum_{n=0}^\infty (-1)^n 2^k c_k \binom{n+k-1}{n} z^{2n+k}.\tag {3.4}$$

Let $m=2n+k$, since $n\in\left[0,\lfloor{m-1\rfloor}/2\right]$, $$h\left(\frac{2z}{1+z^2}\right)=c_0+\sum_{m=1}^\infty \sum_{n=0}^{\lfloor{m-1\rfloor}/2} (-1)^n 2^{m-2n} c_{m-2n} \binom{m-n-1}{n} z^m.\tag {3.5}$$

Claim 1: $$\forall k\in \mathbb N,c_{2k}=0.$$

The statement can be proved by mathematical induction.

Compare constant term on both sides in $(3.1)$, $c_0=2c_0\Rightarrow c_0=0.$

From $(3.5)$, since $c_{2k+2}$ is derived from the linear combination of $c_0,c_2,c_4,\cdots,c_{2k}$, it follows that $c_{2k+2}=0$. Claim 1 is established.

Claim 2: $$\forall k\in \mathbb N,c_{2k+1}=\frac{c_1}{2k+1}.$$

How can Claim 2 be proven?

Fourth attempt

I think of the following equation and try to solve it:

$$g\left(\frac{x+y}{1+xy}\right)=g(x)+g(y).\tag {***}$$

We can arrive $(2.2)$ by setting $x=y=z$.

Let $W(x,y)=\frac{x+y}{1+xy}$, $R(u,v)=u+v$.

Rewrite $(***)$ as

$$g\left(W(x,y)\right)=R\left(g(x),g(y)\right).\tag {4.1}$$

Could $g$ in $(4.1)$ be solved through partial derivatives?

Apart from above attempts, any other thoughts? Thanks.

Given that the limit $c=\lim_{z\rightarrow 0} f(z)$ is assumed to exist (but $f$ is not assumed e.g. analytic at zero) a dynamical systems approach using the $\tanh$ substitution is probably the easiest. You have for $z$ non-zero $$ f \left( \frac{2z}{1+z^2} \right) = 2z\frac{1+z^2}{2z} f(z)$$ Setting $z=\tanh t/2$, $t\in {\Bbb R}\setminus \{0\}$ the above reads: $$f(\tanh(t)) = \frac{2 \tanh(t/2)}{\tanh(t)} f(\tanh (t/2))$$ Iterating this relation $n$ times: $$f(\tanh(t)) = \frac{2^n \tanh(2^{-n} t)}{\tanh(t)} f(\tanh (2^{-n}t))\rightarrow \frac{t }{\tanh t} c$$ as $n\rightarrow \infty$. It follows that $f(\tanh(t))=ct/\tanh(t)$ or $$f(z)=c \frac{{\rm Arctanh} \;z}{z}.$$

Sorry for editing the solution by WORD.