I'm analyzing the following isometric immersion of $\mathbb H^2$ in $\mathbb R^\infty$ given by $f(x,y)=(x_1,x_2,\dots,x_{2m-1},x_{2m},\dots)$ with 
\begin{align}\label{5.1}
    x_{2m-1}=\operatorname{Re}\frac{(x+iy)^m}{\sqrt{m}},\quad x_{2m}=\operatorname{Im}\frac{(x+iy)^m}{\sqrt{m}},\quad m=1,2,\dots
\end{align}
I tried to check that it really is an isometric immersion, but I cannot calculate $f^*g_{\mathbb R^\infty}=g$, some metric $g$, or give it shape, I have tried to do it by means of its polar representation but I have gotten confused without reaching anything concrete. Any ideas how to attack this problem?

[Here][1] I leave the original document.

My attempt was:

>Instead of taking real variable I take complex variable, that is let $z_m=\dfrac{z^m}{\sqrt{m}}$, donde $z_m=x_{2m-1}+ix_{2m}$. Then $dz_m=\sqrt{m}z^{m-1}dz$, thus 
\begin{align*}
\sum_{m=1}^\infty dx_{m}^2&=\sum_{m=1}^\infty (dx_{2m-1}^2+dx_{2m}^2)\\
&=\sum_{m=1}^\infty |dz_m|^2\\
&=\sum_{m=1}^{\infty}m|z|^{2m-2}|dz|^2\\
&=\frac{|dz|^2}{(1-|z|^2)^2}\\[1mm]
&=\frac{dx^2+dy^2}{(1-(x^2+y^2)^2)^2}
\end{align*}
It would only be necessary to prove this metric corresponds to a surface of negative Gaussian curvature.

  [1]: https://www.e-periodica.ch/cntmng?type=pdf&pid=com-001:1932:4::24