# modular function produced via infinite product of weight-k modular forms

Suppose I have a modular form $f_{k}(z)$ of weight $k$:

$f_{k}\left(\frac{az+b}{cz+d}\right) = (cz+d)^{k} f_{k}(z)$

Then consider the following infinite product:

$u(z) = \prod_{n=1}^{\infty} \frac{f_{4n-3}(z)f_{4n}(z)}{f_{4n-2}(z)f_{4n-1}(z)}$

Ignoring convergence problems for the moment, formally $u(z)$ must be a modular function, since:

$\prod_{n=1}^{\infty} \frac{f_{4n-3}(z)f_{4n}(z)}{f_{4n-2}(z)f_{4n-1}(z)} = \prod_{n=1}^{\infty} \frac{(cz+d)^{4n-3}f_{4n-3}(z)(cz+d)^{4n}f_{4n}(z)}{(cz+d)^{4n-2}f_{4n-2}(z)(cz+d)^{4n-1}f_{4n-1}(z)} = \prod_{n=1}^{\infty} \frac{f_{4n-3}\left(\frac{az+b}{cz+d}\right)f_{4n}\left(\frac{az+b}{cz+d}\right)}{f_{4n-2}\left(\frac{az+b}{cz+d}\right)f_{4n-1}\left(\frac{az+b}{cz+d}\right)}$

And
$\frac{(cz+d)^{4n-3}(cz+d)^{4n}}{(cz+d)^{4n-2}(cz+d)^{4n-1}} = (cz+d)^{4n-3+4n-4n+2-4n+1)} = (cz+d)^{0} =1$