The modular discriminant and the j-invariant

The modular discriminant is

$\mathrm{\Delta }(\tau ):=\frac{E_4(\tau {)}^3-E_6(\tau {)}^2}{1728},$

a cusp form of weight $12$ with $q$-expansion $q-24q^2+O(q^3)$ and without zeros on $\mathfrak{H}$ (The discriminant is a nonvanishing cusp form of weight 12, Eisenstein series are modular forms; their Fourier coefficients). The $j$-invariant is the quotient

$j(\tau ):=\frac{E_4(\tau {)}^3}{\mathrm{\Delta }(\tau )},\tau \in \mathfrak{H},$

which is holomorphic on $\mathfrak{H}$ because $\mathrm{\Delta }$ does not vanish there (Meromorphic functions on a plane domain). Since $E_4^3$ and $\mathrm{\Delta }$ both transform with the factor $(c\tau +d{)}^{12}$ under every $\gamma \in S{L}_2(\mathbb{Z})$, the quotient satisfies $j(\gamma \cdot \tau )=j(\tau )$ (Level-one modular forms and cusp forms).

The $q$-expansion and the cusp. From $E_4^3=1+720q+O(q^2)$ and $\mathrm{\Delta }=q(1-24q+O(q^2))$ (Eisenstein series are modular forms; their Fourier coefficients, The discriminant is a nonvanishing cusp form of weight 12),

$j(\tau )=q^{-1}\frac{1+720q+O(q^2)}{1-24q+O(q^2)}=q^{-1}+744+O(q),$

so $j$ has a simple pole at the cusp: its associated function of $q$ is meromorphic at $q=0$ with a pole of order one. The special values follow from the zeros of $E_4,E_6$ (The zeros of E4 and E6 at the elliptic points): $E_6(i)=0$ and $E_4(i)\mathrm{\ne }0$ give $j(i)=\frac{E_4(i{)}^3}{E_4(i{)}^3\mathrm{/}1728}=1728$, while $E_4(\omega )=0$ gives $j(\omega )=0$; the valence formula (The level-one valence formula) is what makes these the only relevant zeros.