I have been trying to find an analytic onto function from $\mathbb{C} \to D$ where $D$ is open unit disc. I had a few candidate functions in mind, but couldn't prove that they are analytic due to a $\frac{\bar{h}}{h}$ term remaining which has no limit. On the Wikipedia page of Unit Disc, it states- "The function $f(z)={\frac {z}{1-|z|^{2}}}$ is an example of a real analytic and bijective function from the open unit disk to the plane; its inverse function is also analytic."
I tried to prove its analyticity, but encountered the same problem. Here is my attempt- $$\lim\limits_{h \to 0} \frac{f(z+h) - f(z)}{h} \\ \implies \lim\limits_{h \to 0} \frac{\frac{z+h}{1-|z+h|^2} - \frac{z}{1-|z|^2}}{h} \\ \implies \lim\limits_{h \to 0} \frac{z+h - z|z|^2 -h|z|^2 - z + z|z+h|^2}{h(1-|z+h|^2)(1-|z|^2)} \\ \implies \lim\limits_{h \to 0} \frac{h - z|z|^2 -h|z|^2 + z|z|^2+z|h|^2 + z\bar{z}h + z^2\bar{h}}{h(1-|z+h|^2)(1-|z|^2)} \\ \implies \lim\limits_{h \to 0} \frac{h -h|z|^2 +z(h\bar{h})^2 + |z|h + z^2\bar{h}}{h(1-|z+h|^2)(1-|z|^2)}$$
Now, except $z^2\bar{h}$, all terms are divisible h. This makes me think the limit does not exist and the function is not differentiable but its stated in the link that it is analytic. How do I proceed?