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 $f^*g_{\mathbb R^\infty}=\displaystyle\sum_{m=1}^\infty dx_m^2$ 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 I leave the original document.

My attempt was (I know it is wrong but I tried to get to what is mentioned in the document, unfortunately I have not arrived):

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}mz^{2m-2}dz^2\\ &=\frac{dz^2}{z^2}\sum_{m=1}^{\infty}mz^{2m}\\ &=\frac{dz^2}{z^2(1-z^2)^2} \end{align*}