Question:
Exercise 1.1.8 Recall that $\mu_3$ denotes the complex cube root of unity $e^{2\pi i/3}$. Show that $$\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix} \mu_3 = \mu_3 +1$$ and by periodicity $$g_2(\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix} \mu_3) = g_2(\mu_3).$$ Show that by modularity also $$g_2(\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix} \mu_3) = \mu_3^4 g_2(\mu_3)$$ and therefore $g_2(\mu_3)=0$. Conclude that $g_3(\mu_3) \neq 0$ and $j(\mu_3) =0$. Argue similarly to show that $g_3(i) =0$, $g_2(i) \neq 0$ and $j(i) = 1728$.
My attempt: I am able to show that $g_2(\mu_3) =0$ and $g_3(i) =0$. However, is there any way I can show that $g_3(\mu_3) \neq 0$ and $g_2(i) \neq 0$ without the use of the valence formula? Any hints are appreciated.
Source: Diamond and Shurman's A Fist Course in Modular Forms