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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?

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

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