Mixing
Möbius Transformation Example
$begingroup$
I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?
complex-analysis
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
$endgroup$
add a comment |
$begingroup$
I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?
complex-analysis
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
$endgroup$
add a comment |
$begingroup$
I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?
complex-analysis
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
$endgroup$
I'm interested in the form of the Möbius transformation of the following mappings: $2i rightarrow 0, -1 rightarrow infty, 0 rightarrow -4i$. When I tried to solve the system of equations I had trouble to interpret the mapping $-1 rightarrow infty$. Could anybody give me some advice?
complex-analysis
complex-analysis
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
asked Nov 9 '18 at 13:33
John SmithJohn Smith
538
538
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Is $z mapsto frac{az+b}{z+1}$
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
$endgroup$
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
1
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
1
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
add a comment |
$begingroup$
A mobius transformation may be given as a function
$$
f(z) = frac{az+b}{cz + d}
$$
with some niceness restrictions on $a, b, c, d$.
If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
$endgroup$
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
add a comment |
$begingroup$
When working with a Möbius transformation
$$f(x) = frac{ax+b}{cx+d}
$$
you are allowed to do just a little bit of "infinity arithmetic". For example:
$$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
$$
which equals $infty$ when $c=0$.
And then a simple computation gives
$$f^{-1}(x) = frac{da - b}{-cx+a}
$$
and so
$$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
$$
which, again, equals $infty$ when $c=0$.
You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
$endgroup$
add a comment |
$begingroup$
I come up with an alternate answer where you can easily get rid of the limits.
Let the transformation be
$w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:
$0=frac{2ai+b}{2ci+d},
infty =frac{-a+b}{-c+d} $ ..We get $-c+d =0$
And the equation is $b=-4id$
Now solve for $a,b,c,d$. We get
$a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
Put these values of $a,b,c,d$ in the transformation and simplify we get
$w=frac{z+2/i}{z/2+1/2} $.
And hence $w= frac{2z-4i}{z+1}$.
edited Jan 9 at 15:43
Thomas Shelby
2,877421
answered Nov 9 '18 at 14:20
HenryHenry
327
$endgroup$
1
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Is $z mapsto frac{az+b}{z+1}$
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
$endgroup$
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
1
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
1
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
add a comment |
$begingroup$
Is $z mapsto frac{az+b}{z+1}$
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
$endgroup$
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
1
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
1
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
add a comment |
$begingroup$
Is $z mapsto frac{az+b}{z+1}$
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
$endgroup$
Is $z mapsto frac{az+b}{z+1}$
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
answered Nov 9 '18 at 13:37
Richard MartinRichard Martin
1,61318
1,61318
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
1
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
1
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
add a comment |
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
1
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
1
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
$begingroup$
Still don't get it … then $az+b = -4i$, therefore $-4ia = b$. Then I have to eliminate $a$ somehow.
$endgroup$
– John Smith
Nov 9 '18 at 13:40
1
1
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
Use the fact that $2i$ maps to $0$, and $0$ to $-4i$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:44
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
$begingroup$
So then $frac{2ai-4ai}{2i+1}=0$ and therefore $2ai-4ai = -2ai = 0$, and then $a=0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:47
1
1
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
No, because your reasoning $az+b=-4i$ is wrong: it's only true when $z=0$. The correct answer is $b=-4i$, and $a=2$.
$endgroup$
– Richard Martin
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
$begingroup$
Ah ups. Okay thank you!
$endgroup$
– John Smith
Nov 9 '18 at 13:50
add a comment |
$begingroup$
A mobius transformation may be given as a function
$$
f(z) = frac{az+b}{cz + d}
$$
with some niceness restrictions on $a, b, c, d$.
If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
$endgroup$
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
add a comment |
$begingroup$
A mobius transformation may be given as a function
$$
f(z) = frac{az+b}{cz + d}
$$
with some niceness restrictions on $a, b, c, d$.
If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
$endgroup$
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
add a comment |
$begingroup$
A mobius transformation may be given as a function
$$
f(z) = frac{az+b}{cz + d}
$$
with some niceness restrictions on $a, b, c, d$.
If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
$endgroup$
A mobius transformation may be given as a function
$$
f(z) = frac{az+b}{cz + d}
$$
with some niceness restrictions on $a, b, c, d$.
If we have $f(z_0) = infty$, that means that $z_0$ is a root of the denominator. In other words, $cz_0 + d = 0$. In your case, that means that you want $ccdot (-1) + d = 0$.
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
answered Nov 9 '18 at 13:39
ArthurArthur
114k7115197
114k7115197
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
add a comment |
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
$begingroup$
Does this imply that $a = 0$?
$endgroup$
– John Smith
Nov 9 '18 at 13:43
add a comment |
$begingroup$
When working with a Möbius transformation
$$f(x) = frac{ax+b}{cx+d}
$$
you are allowed to do just a little bit of "infinity arithmetic". For example:
$$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
$$
which equals $infty$ when $c=0$.
And then a simple computation gives
$$f^{-1}(x) = frac{da - b}{-cx+a}
$$
and so
$$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
$$
which, again, equals $infty$ when $c=0$.
You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
$endgroup$
add a comment |
$begingroup$
When working with a Möbius transformation
$$f(x) = frac{ax+b}{cx+d}
$$
you are allowed to do just a little bit of "infinity arithmetic". For example:
$$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
$$
which equals $infty$ when $c=0$.
And then a simple computation gives
$$f^{-1}(x) = frac{da - b}{-cx+a}
$$
and so
$$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
$$
which, again, equals $infty$ when $c=0$.
You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
$endgroup$
add a comment |
$begingroup$
When working with a Möbius transformation
$$f(x) = frac{ax+b}{cx+d}
$$
you are allowed to do just a little bit of "infinity arithmetic". For example:
$$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
$$
which equals $infty$ when $c=0$.
And then a simple computation gives
$$f^{-1}(x) = frac{da - b}{-cx+a}
$$
and so
$$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
$$
which, again, equals $infty$ when $c=0$.
You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
$endgroup$
When working with a Möbius transformation
$$f(x) = frac{ax+b}{cx+d}
$$
you are allowed to do just a little bit of "infinity arithmetic". For example:
$$frac{a cdot infty + b}{c cdot infty + d}=frac{a}{c}
$$
which equals $infty$ when $c=0$.
And then a simple computation gives
$$f^{-1}(x) = frac{da - b}{-cx+a}
$$
and so
$$f^{-1}(infty) = frac{d cdot infty - b}{-c cdot infty + a} = frac{d}{-c}
$$
which, again, equals $infty$ when $c=0$.
You can, of course, justify this bit of "infinity arithmetic" rigorously using limits.
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
answered Nov 9 '18 at 13:45
Lee MosherLee Mosher
49k33685
49k33685
add a comment |
add a comment |
$begingroup$
I come up with an alternate answer where you can easily get rid of the limits.
Let the transformation be
$w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:
$0=frac{2ai+b}{2ci+d},
infty =frac{-a+b}{-c+d} $ ..We get $-c+d =0$
And the equation is $b=-4id$
Now solve for $a,b,c,d$. We get
$a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
Put these values of $a,b,c,d$ in the transformation and simplify we get
$w=frac{z+2/i}{z/2+1/2} $.
And hence $w= frac{2z-4i}{z+1}$.
edited Jan 9 at 15:43
Thomas Shelby
2,877421
answered Nov 9 '18 at 14:20
HenryHenry
327
$endgroup$
1
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
add a comment |
$begingroup$
I come up with an alternate answer where you can easily get rid of the limits.
Let the transformation be
$w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:
$0=frac{2ai+b}{2ci+d},
infty =frac{-a+b}{-c+d} $ ..We get $-c+d =0$
And the equation is $b=-4id$
Now solve for $a,b,c,d$. We get
$a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
Put these values of $a,b,c,d$ in the transformation and simplify we get
$w=frac{z+2/i}{z/2+1/2} $.
And hence $w= frac{2z-4i}{z+1}$.
edited Jan 9 at 15:43
Thomas Shelby
2,877421
answered Nov 9 '18 at 14:20
HenryHenry
327
$endgroup$
1
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
add a comment |
$begingroup$
I come up with an alternate answer where you can easily get rid of the limits.
Let the transformation be
$w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:
$0=frac{2ai+b}{2ci+d},
infty =frac{-a+b}{-c+d} $ ..We get $-c+d =0$
And the equation is $b=-4id$
Now solve for $a,b,c,d$. We get
$a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
Put these values of $a,b,c,d$ in the transformation and simplify we get
$w=frac{z+2/i}{z/2+1/2} $.
And hence $w= frac{2z-4i}{z+1}$.
edited Jan 9 at 15:43
Thomas Shelby
2,877421
answered Nov 9 '18 at 14:20
HenryHenry
327
$endgroup$
I come up with an alternate answer where you can easily get rid of the limits.
Let the transformation be
$w= frac{az+b}{cz+d} $. Now substitute the corresponding values of $z$ and $w$. Now you will get three equation as follows:
$0=frac{2ai+b}{2ci+d},
infty =frac{-a+b}{-c+d} $ ..We get $-c+d =0$
And the equation is $b=-4id$
Now solve for $a,b,c,d$. We get
$a=frac{b}{(2/i)}= frac{c}{(1/2)}=frac{d}{(1/2)} $.
Put these values of $a,b,c,d$ in the transformation and simplify we get
$w=frac{z+2/i}{z/2+1/2} $.
And hence $w= frac{2z-4i}{z+1}$.
edited Jan 9 at 15:43
Thomas Shelby
2,877421
answered Nov 9 '18 at 14:20
HenryHenry
327
edited Jan 9 at 15:43
Thomas Shelby
2,877421
edited Jan 9 at 15:43
Thomas Shelby
2,877421
edited Jan 9 at 15:43
Thomas Shelby
2,877421
2,877421
answered Nov 9 '18 at 14:20
HenryHenry
327
answered Nov 9 '18 at 14:20
HenryHenry
327
answered Nov 9 '18 at 14:20
HenryHenry
327
327
1
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
add a comment |
1
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
1
1
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
And if u encounter with infinity jus don't worry ..use the concept of limits as we do in calculus replace infinity wid n and then take the limit n tends to infinity ...I hope it will help
$endgroup$
– Henry
Nov 9 '18 at 14:22
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
$begingroup$
Very good advice!
$endgroup$
– John Smith
Nov 9 '18 at 14:25
add a comment |
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