I have to show that there is a Cusp Form $f \in S_{12n}(SL_2(\mathbb{Z}))$ such that $f(a_i)=b_i$ for $1\leq i\leq n$ with $a_i$ in the fundamental domain and $b_i \in \mathbb{C}$.
I tried to get some intuition of the proof by looking at $n=1$ so $f$ has to be a multiple of $\Delta$. I tried to find the factor such that $\Delta(2i)=0$. So my idea was to shift the root in $\infty$ to any point in the fundamental domain. However, I have no idea how to do it. Since the assumption says that $a_i$ has to be in the fundamental domain I think one has to use Mobius transformations. Thanks for your help.
Folow-up question: Is $f$ unique?
I presume the $a_i$ are distinct points in the fundamental domain excluding the point at infinity. The dimension of this space $S=S_{12n}(\text{SL}_2(\Bbb Z))$ of cusp forms is $n$. We are looking at the map $\Phi:S\to\Bbb C^n$ given by $\Phi(f)=(f(a_1),\ldots,f(a_n))$ and wanting to prove it surjective. As both these vector spaces have dimension $n$ it suffices to prove $\Phi$ injective. A nonzero element in the kernel of $\Phi$ will have $n+1$ zeroes in the fundamental domain including the zero at infinity. This is too many.
There are slight problems with this argument if the $a_i$ include points equivalent to $i$ or $\frac12(1+i\sqrt3)$. You have to show that modular forms of weight $12n$ have even order at $i$ and order divisible by $3$ at $\frac12(1+i\sqrt3)$, but this follows from the zero-counting formula for modular forms.