# When an 'integer-complex' always is transformed to an 'integer-complex'?

I tried Mobius transformations $$\frac{az+b}{cz+d}$$ for several $$z∈ \mathbb{Z}+i\mathbb{Z}$$ and it seems that even if $$M=\begin{pmatrix}a&b\\c&d\end{pmatrix} \in SL_2(\mathbb{Z})$$, not any 'integer-complex' ($$z∈ \mathbb{Z}+i\mathbb{Z}$$) transforms to a 'integer-complex' and vice-versa. If I am not mistaken after evaluation of $$\frac{az+b}{cz+d}$$ even lines can be transformed to circles and and vice-versa.

My question is, for what subset of $$SL_2(\mathbb{Z})$$, a $$z∈ \mathbb{Z}+i\mathbb{Z}$$ always is transformed to a $$w∈ \mathbb{Z}+i\mathbb{Z}$$ by transformations $$\frac{az+b}{cz+d}$$? In other words, a set that transforms lattices to lattices?

• For the record, the elements of the set $\mathbf{Z}[i]:=\mathbf{Z}+i\mathbf{Z}$ are called Gaussian integers. – Keenan Kidwell Dec 25 '18 at 19:14

Multiply nominator and denominator of $$\frac{az+b}{cz+d}$$ by $$\overline{cz+d}$$ for $$a,b,c,d\in \Bbb{Z}$$ to see that we necessarily need $$\frac{ad-bc}{c^2+d^2}=\frac{1}{c^2+d^2}\in \Bbb{Z}.$$ So we need $$c^2+d^2=1$$ as one necessary condition. Now you can finish.