Mixing
Mobius transform |z|<1 to the right half plane
$begingroup$
Find a Mobius transformation mapping the unit disk {|z| < 1} into the right half-plane and taking z = −i to the origin.
My workings:
$phi(t) = frac{az+b}{cz+d}$
We map -i to the origin (0) by taking the numerator and equating it to 0, knowing b = +i (since we started at -i).
That is:
$az+i=0 implies i = b, a = 1$, which is correct.
However, how do we map $|z|$ to the upper right half plane (i.e. Re(w) = 0)?
Thanks.
complex-analysis mobius-transformation
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
$endgroup$
add a comment |
$begingroup$
Find a Mobius transformation mapping the unit disk {|z| < 1} into the right half-plane and taking z = −i to the origin.
My workings:
$phi(t) = frac{az+b}{cz+d}$
We map -i to the origin (0) by taking the numerator and equating it to 0, knowing b = +i (since we started at -i).
That is:
$az+i=0 implies i = b, a = 1$, which is correct.
However, how do we map $|z|$ to the upper right half plane (i.e. Re(w) = 0)?
Thanks.
complex-analysis mobius-transformation
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
$endgroup$
add a comment |
$begingroup$
Find a Mobius transformation mapping the unit disk {|z| < 1} into the right half-plane and taking z = −i to the origin.
My workings:
$phi(t) = frac{az+b}{cz+d}$
We map -i to the origin (0) by taking the numerator and equating it to 0, knowing b = +i (since we started at -i).
That is:
$az+i=0 implies i = b, a = 1$, which is correct.
However, how do we map $|z|$ to the upper right half plane (i.e. Re(w) = 0)?
Thanks.
complex-analysis mobius-transformation
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
$endgroup$
Find a Mobius transformation mapping the unit disk {|z| < 1} into the right half-plane and taking z = −i to the origin.
My workings:
$phi(t) = frac{az+b}{cz+d}$
We map -i to the origin (0) by taking the numerator and equating it to 0, knowing b = +i (since we started at -i).
That is:
$az+i=0 implies i = b, a = 1$, which is correct.
However, how do we map $|z|$ to the upper right half plane (i.e. Re(w) = 0)?
Thanks.
complex-analysis mobius-transformation
complex-analysis mobius-transformation
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
edited Jan 27 at 6:37
Kavi Rama Murthy
69.4k53170
69.4k53170
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
asked Jan 27 at 4:52
Dr.DoofusDr.Doofus
12612
12612
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Watch that first step: $-ai+b=0implies b=ai$. Let's have $-1to i$, so that $-a+ai=i(-c+d)$. And $0to1$, giving $b=d$. So $c=-a$.
So, we get $f(z)=frac{az+ai}{-az+ai}$ or $boxed{f(z)=frac{z+i}{i-z}}$.
By specifying the values at $3$ points, the Möbius transformation is determined.
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
1
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
add a comment |
$begingroup$
It is $frac {1-iz} {1+iz}$. Multiply numerator and denominator by $1-ioverline {z}$ to prove that this works.
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
1
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Watch that first step: $-ai+b=0implies b=ai$. Let's have $-1to i$, so that $-a+ai=i(-c+d)$. And $0to1$, giving $b=d$. So $c=-a$.
So, we get $f(z)=frac{az+ai}{-az+ai}$ or $boxed{f(z)=frac{z+i}{i-z}}$.
By specifying the values at $3$ points, the Möbius transformation is determined.
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
1
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
add a comment |
$begingroup$
Watch that first step: $-ai+b=0implies b=ai$. Let's have $-1to i$, so that $-a+ai=i(-c+d)$. And $0to1$, giving $b=d$. So $c=-a$.
So, we get $f(z)=frac{az+ai}{-az+ai}$ or $boxed{f(z)=frac{z+i}{i-z}}$.
By specifying the values at $3$ points, the Möbius transformation is determined.
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
$endgroup$
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
1
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
add a comment |
$begingroup$
Watch that first step: $-ai+b=0implies b=ai$. Let's have $-1to i$, so that $-a+ai=i(-c+d)$. And $0to1$, giving $b=d$. So $c=-a$.
So, we get $f(z)=frac{az+ai}{-az+ai}$ or $boxed{f(z)=frac{z+i}{i-z}}$.
By specifying the values at $3$ points, the Möbius transformation is determined.
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
$endgroup$
Watch that first step: $-ai+b=0implies b=ai$. Let's have $-1to i$, so that $-a+ai=i(-c+d)$. And $0to1$, giving $b=d$. So $c=-a$.
So, we get $f(z)=frac{az+ai}{-az+ai}$ or $boxed{f(z)=frac{z+i}{i-z}}$.
By specifying the values at $3$ points, the Möbius transformation is determined.
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 6:41
Chris CusterChris Custer
14.2k3827
14.2k3827
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
1
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
add a comment |
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
1
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
$begingroup$
Correct answer. Can you elaborate on why you said "let's have -1 -> i? And how did you choose -1?
$endgroup$
– Dr.Doofus
Jan 27 at 7:31
1
1
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
$begingroup$
$-1$ is another point on $mid zmid=1$ (besides $-i$), and I wish to move it onto the $y$-axis.
$endgroup$
– Chris Custer
Jan 27 at 7:34
add a comment |
$begingroup$
It is $frac {1-iz} {1+iz}$. Multiply numerator and denominator by $1-ioverline {z}$ to prove that this works.
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
1
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
add a comment |
$begingroup$
It is $frac {1-iz} {1+iz}$. Multiply numerator and denominator by $1-ioverline {z}$ to prove that this works.
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
1
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
add a comment |
$begingroup$
It is $frac {1-iz} {1+iz}$. Multiply numerator and denominator by $1-ioverline {z}$ to prove that this works.
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
$endgroup$
It is $frac {1-iz} {1+iz}$. Multiply numerator and denominator by $1-ioverline {z}$ to prove that this works.
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
answered Jan 27 at 5:09
Kavi Rama MurthyKavi Rama Murthy
69.4k53170
69.4k53170
1
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
add a comment |
1
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
1
1
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
You said upper half plane in the title and right half plane in the question. I have taken it as right half plane.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 5:11
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
Yes, thanks for spotting that. I've edited the question title. In regards to your answer, how do you get that value?
$endgroup$
– Dr.Doofus
Jan 27 at 6:09
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
$begingroup$
You can start with $1-iz$ in the numerator so that it vanishes at $-i$. Take denominator as $a+bz$, multiply numerator and denominator by the conjugate of the denominator. You then want the real part of the numerator to be positive when $|z| <1$. It should be easy to guess what $a$ and $b$ should be.
$endgroup$
– Kavi Rama Murthy
Jan 27 at 6:36
add a comment |
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