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












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I tried Mobius transformations $frac{az+b}{cz+d}$ for several $z∈ mathbb{Z}+imathbb{Z}$ and it seems that even if $M=begin{pmatrix}a&b\c&dend{pmatrix} in SL_2(mathbb{Z})$, not any 'integer-complex' ($z∈ mathbb{Z}+imathbb{Z}$) 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 $z∈ mathbb{Z}+imathbb{Z}$ always is transformed to a $w∈ mathbb{Z}+imathbb{Z}$ by transformations $frac{az+b}{cz+d}$? In other words, a set that transforms lattices to lattices?










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  • 2




    For the record, the elements of the set $mathbf{Z}[i]:=mathbf{Z}+imathbf{Z}$ are called Gaussian integers.
    – Keenan Kidwell
    2 days ago
















0














I tried Mobius transformations $frac{az+b}{cz+d}$ for several $z∈ mathbb{Z}+imathbb{Z}$ and it seems that even if $M=begin{pmatrix}a&b\c&dend{pmatrix} in SL_2(mathbb{Z})$, not any 'integer-complex' ($z∈ mathbb{Z}+imathbb{Z}$) 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 $z∈ mathbb{Z}+imathbb{Z}$ always is transformed to a $w∈ mathbb{Z}+imathbb{Z}$ by transformations $frac{az+b}{cz+d}$? In other words, a set that transforms lattices to lattices?










share|cite|improve this question




















  • 2




    For the record, the elements of the set $mathbf{Z}[i]:=mathbf{Z}+imathbf{Z}$ are called Gaussian integers.
    – Keenan Kidwell
    2 days ago














0












0








0







I tried Mobius transformations $frac{az+b}{cz+d}$ for several $z∈ mathbb{Z}+imathbb{Z}$ and it seems that even if $M=begin{pmatrix}a&b\c&dend{pmatrix} in SL_2(mathbb{Z})$, not any 'integer-complex' ($z∈ mathbb{Z}+imathbb{Z}$) 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 $z∈ mathbb{Z}+imathbb{Z}$ always is transformed to a $w∈ mathbb{Z}+imathbb{Z}$ by transformations $frac{az+b}{cz+d}$? In other words, a set that transforms lattices to lattices?










share|cite|improve this question















I tried Mobius transformations $frac{az+b}{cz+d}$ for several $z∈ mathbb{Z}+imathbb{Z}$ and it seems that even if $M=begin{pmatrix}a&b\c&dend{pmatrix} in SL_2(mathbb{Z})$, not any 'integer-complex' ($z∈ mathbb{Z}+imathbb{Z}$) 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 $z∈ mathbb{Z}+imathbb{Z}$ always is transformed to a $w∈ mathbb{Z}+imathbb{Z}$ by transformations $frac{az+b}{cz+d}$? In other words, a set that transforms lattices to lattices?







complex-analysis






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edited 2 days ago

























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72D

547116




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  • 2




    For the record, the elements of the set $mathbf{Z}[i]:=mathbf{Z}+imathbf{Z}$ are called Gaussian integers.
    – Keenan Kidwell
    2 days ago














  • 2




    For the record, the elements of the set $mathbf{Z}[i]:=mathbf{Z}+imathbf{Z}$ are called Gaussian integers.
    – Keenan Kidwell
    2 days ago








2




2




For the record, the elements of the set $mathbf{Z}[i]:=mathbf{Z}+imathbf{Z}$ are called Gaussian integers.
– Keenan Kidwell
2 days ago




For the record, the elements of the set $mathbf{Z}[i]:=mathbf{Z}+imathbf{Z}$ are called Gaussian integers.
– Keenan Kidwell
2 days ago










1 Answer
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Multiply nominator and denominator of $frac{az+b}{cz+d}$ by $overline{cz+d}$ for $a,b,c,din Bbb{Z}$ 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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    1 Answer
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    Multiply nominator and denominator of $frac{az+b}{cz+d}$ by $overline{cz+d}$ for $a,b,c,din Bbb{Z}$ 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.






    share|cite|improve this answer


























      1














      Multiply nominator and denominator of $frac{az+b}{cz+d}$ by $overline{cz+d}$ for $a,b,c,din Bbb{Z}$ 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.






      share|cite|improve this answer
























        1












        1








        1






        Multiply nominator and denominator of $frac{az+b}{cz+d}$ by $overline{cz+d}$ for $a,b,c,din Bbb{Z}$ 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.






        share|cite|improve this answer












        Multiply nominator and denominator of $frac{az+b}{cz+d}$ by $overline{cz+d}$ for $a,b,c,din Bbb{Z}$ 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Dietrich Burde

        77.4k64386




        77.4k64386






























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