When an 'integer-complex' always is transformed to an 'integer-complex'?
I tried Mobius transformations $frac{az+b}{cz+d}$ for several $z∈ mathbb{Z}+imathbb{Z}$ 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
asked 2 days ago
72D
547116
add a comment |
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
asked 2 days ago
72D
547116
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
add a comment |
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
asked 2 days ago
72D
547116
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
complex-analysis
asked 2 days ago
72D
547116
asked 2 days ago
72D
547116
edited 2 days ago
asked 2 days ago
72D
547116
asked 2 days ago
72D
547116
asked 2 days ago
72D
547116
547116
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
add a comment |
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
add a comment |
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.
answered 2 days ago
Dietrich Burde
77.4k64386
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 2 days ago
Dietrich Burde
77.4k64386
add a comment |
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.
answered 2 days ago
Dietrich Burde
77.4k64386
add a comment |
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.
answered 2 days ago
Dietrich Burde
77.4k64386
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.
answered 2 days ago
Dietrich Burde
77.4k64386
answered 2 days ago
Dietrich Burde
77.4k64386
answered 2 days ago
Dietrich Burde
77.4k64386
answered 2 days ago
Dietrich Burde
77.4k64386
77.4k64386
add a comment |
add a comment |
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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