Generating the special linear group of 2 by 2 matrices over the integers. Our Number Theory professor claimed that the special linear group $\text{SL}_2(\mathbb{Z})$ is generated by just two matrices: 
$$
M_1=\begin{pmatrix} 0& -1\\ 1& 0 \\\end{pmatrix} 
$$
$$
M_2=\begin{pmatrix} 1& 1\\ 1& 0 \\\end{pmatrix}
$$
He commented that the proof is outside the scope of the syllabus. When I pressed him to give a hint, he made the following cryptic comment:

Consider the action of the group $\text{SL}_2(\mathbb{Z})$ on the complex upper half plane $\mathbb{H}$ by the action:
  $$\begin{pmatrix} a& b\\ c& d \\\end{pmatrix}: z \to \frac{az+b}{cz+d}$$
  Show that 'some region' in $\mathbb{H}$ is a 'fundamental domain' and that finishes the proof. 

I don't follow the argument at all. He drew a region that looked like a semicircle with a rectangular patch in between. This is the 'some region' in the block quote. 
I don't know how the complex plane helps. I don't know what a 'fundamental domain' means in this problem. And most of all, I don't see how this approaches the proof at all :(
Would someone explain this argument to me? I will be happy with references, papers or a general hand waving argument as well.
Thank you.

Yes, the claim is correct. The proof uses a construction of a fundamental modular domain for the modular group $SL_2(\mathbb{Z})$.
It is very well explained in the notes of Keith Conrad here; see Theorem $1.1$. The proof there is very clear and detailed. It goes as your professor has hinted. In fact, Conrad gives two proofs. First an algebraic one, and then a geometric one, with the fundamental domain.
The elements of the modular group can be viewed as  linear fractional transformations of the upper half of the complex plane. This helps a lot to study this group. For more details see the article on the modular group.
