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How to find transformation from the upper half plane into the right half plane?
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Find the general form of the linear transformation which transforms
the upper half plane into the right half plane.
In my notes I have a Mobius transformation from the upper half plane to the unit circle $T(z)=e^itheta_0fracz-z_0z-overline z_0$.
Also another transformation from the unit circle to the upper half plane $T(z)=(1-i)fracz-iz-1$.
But I do not know how to construct the possible composition transformation from upper half plane into the right half plane.
Could someone help please?
Any hint?
complex-analysis linear-transformations mobius-transformation
asked yesterday
Isa
327217
 |Â
show 8 more comments
up vote
0
down vote
favorite
Find the general form of the linear transformation which transforms
the upper half plane into the right half plane.
In my notes I have a Mobius transformation from the upper half plane to the unit circle $T(z)=e^itheta_0fracz-z_0z-overline z_0$.
Also another transformation from the unit circle to the upper half plane $T(z)=(1-i)fracz-iz-1$.
But I do not know how to construct the possible composition transformation from upper half plane into the right half plane.
Could someone help please?
Any hint?
complex-analysis linear-transformations mobius-transformation
asked yesterday
Isa
327217
6
How about you just rotate the plane?
– quid♦
yesterday
@quid Ok, Am I going to use this transformation $T(z)=e^itheta_0z$ then?.
– Isa
yesterday
Yes. And what is $theta_0$?
– David G. Stork
yesterday
@DavidG.Stork it's the angle of rotation, the principal argument
– Isa
yesterday
1
I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) =az+bover cz+d$$ where $ad-bcneq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $cneq0$ case or to prove that $c=0.$
– saulspatz
yesterday
 |Â
show 8 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the general form of the linear transformation which transforms
the upper half plane into the right half plane.
In my notes I have a Mobius transformation from the upper half plane to the unit circle $T(z)=e^itheta_0fracz-z_0z-overline z_0$.
Also another transformation from the unit circle to the upper half plane $T(z)=(1-i)fracz-iz-1$.
But I do not know how to construct the possible composition transformation from upper half plane into the right half plane.
Could someone help please?
Any hint?
complex-analysis linear-transformations mobius-transformation
asked yesterday
Isa
327217
Find the general form of the linear transformation which transforms
the upper half plane into the right half plane.
In my notes I have a Mobius transformation from the upper half plane to the unit circle $T(z)=e^itheta_0fracz-z_0z-overline z_0$.
Also another transformation from the unit circle to the upper half plane $T(z)=(1-i)fracz-iz-1$.
But I do not know how to construct the possible composition transformation from upper half plane into the right half plane.
Could someone help please?
Any hint?
complex-analysis linear-transformations mobius-transformation
asked yesterday
Isa
327217
asked yesterday
Isa
327217
asked yesterday
Isa
327217
asked yesterday
Isa
327217
327217
6
How about you just rotate the plane?
– quid♦
yesterday
@quid Ok, Am I going to use this transformation $T(z)=e^itheta_0z$ then?.
– Isa
yesterday
Yes. And what is $theta_0$?
– David G. Stork
yesterday
@DavidG.Stork it's the angle of rotation, the principal argument
– Isa
yesterday
1
I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) =az+bover cz+d$$ where $ad-bcneq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $cneq0$ case or to prove that $c=0.$
– saulspatz
yesterday
 |Â
show 8 more comments
6
How about you just rotate the plane?
– quid♦
yesterday
@quid Ok, Am I going to use this transformation $T(z)=e^itheta_0z$ then?.
– Isa
yesterday
Yes. And what is $theta_0$?
– David G. Stork
yesterday
@DavidG.Stork it's the angle of rotation, the principal argument
– Isa
yesterday
1
I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) =az+bover cz+d$$ where $ad-bcneq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $cneq0$ case or to prove that $c=0.$
– saulspatz
yesterday
6
6
How about you just rotate the plane?
– quid♦
yesterday
How about you just rotate the plane?
– quid♦
yesterday
@quid Ok, Am I going to use this transformation $T(z)=e^itheta_0z$ then?.
– Isa
yesterday
@quid Ok, Am I going to use this transformation $T(z)=e^itheta_0z$ then?.
– Isa
yesterday
Yes. And what is $theta_0$?
– David G. Stork
yesterday
Yes. And what is $theta_0$?
– David G. Stork
yesterday
@DavidG.Stork it's the angle of rotation, the principal argument
– Isa
yesterday
@DavidG.Stork it's the angle of rotation, the principal argument
– Isa
yesterday
1
1
I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) =az+bover cz+d$$ where $ad-bcneq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $cneq0$ case or to prove that $c=0.$
– saulspatz
yesterday
I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) =az+bover cz+d$$ where $ad-bcneq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $cneq0$ case or to prove that $c=0.$
– saulspatz
yesterday
 |Â
show 8 more comments
1 Answer
1
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up vote
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I think I've got it. The transformation $T$ must take the boundary of the upper half-plane to the boundary of the right half-plane. That is, it must take the real axis to the imaginary axis. If $$T(z)=az+bover cz +d, ad-bcneq=0,tag1$$ we have that $xinmathbfR$ implies $$Refrac(ax+b)overline(cx+d)cx+d=0impliesRe(ax+b)overline(cx+d)=0$$
Now for given, $a,b,c,dinmathbfC, Re(ax+b)overline(cx+d)$ is a quadratic in $x$ that vanishes everywhere, so all the coefficients must be $0$. That is,
$$Re(aoverlinec)=Re(aoverlined+boverlinec)=Re(boverlined)=0tag2$$
Let us assume that $cneq0.$ Then we may divide numerator and denominator in $(1)$ by $c$, or what is the same thing, we may assume that $c=1,$ so $(2)$ becomes
$$Re(a)=Re(aoverlined+b)=Re(boverlined)=0tag3$$
From $c=1$ we have $T(-d)=infty,$ but $infty$ is on the imaginary axis so $dinmathbfR,$ and from $(3),$ we have $Re a=0$ and $Re(ad+b)=0,$ so that $Re b = 0.$ That is,
$$T(z) = ifracalpha z+betaz+d, text where alpha,beta,dinmathbfR, alpha d -betaneq0$$
which can obviously be re-written more symmetrically as
$$T(z) = ifracalpha z+betagamma z+ delta, text where alpha,beta,gamma deltainmathbfR, alphadelta-betagammaneq=0$$
However, this leaves open the possibility that $T$ maps the upper half-plane to the $left$ half-plane.
This leaves you with two things to do. First, finish off the $cneq0$ case. (Hint: $Re T(i)>0$.) Second, do the (easier) $c=0$ case.
I feel that there must be an easier way of seeing this, but I've not been able to find one.
answered 21 hours ago
saulspatz
10.2k21323
1
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I think I've got it. The transformation $T$ must take the boundary of the upper half-plane to the boundary of the right half-plane. That is, it must take the real axis to the imaginary axis. If $$T(z)=az+bover cz +d, ad-bcneq=0,tag1$$ we have that $xinmathbfR$ implies $$Refrac(ax+b)overline(cx+d)cx+d=0impliesRe(ax+b)overline(cx+d)=0$$
Now for given, $a,b,c,dinmathbfC, Re(ax+b)overline(cx+d)$ is a quadratic in $x$ that vanishes everywhere, so all the coefficients must be $0$. That is,
$$Re(aoverlinec)=Re(aoverlined+boverlinec)=Re(boverlined)=0tag2$$
Let us assume that $cneq0.$ Then we may divide numerator and denominator in $(1)$ by $c$, or what is the same thing, we may assume that $c=1,$ so $(2)$ becomes
$$Re(a)=Re(aoverlined+b)=Re(boverlined)=0tag3$$
From $c=1$ we have $T(-d)=infty,$ but $infty$ is on the imaginary axis so $dinmathbfR,$ and from $(3),$ we have $Re a=0$ and $Re(ad+b)=0,$ so that $Re b = 0.$ That is,
$$T(z) = ifracalpha z+betaz+d, text where alpha,beta,dinmathbfR, alpha d -betaneq0$$
which can obviously be re-written more symmetrically as
$$T(z) = ifracalpha z+betagamma z+ delta, text where alpha,beta,gamma deltainmathbfR, alphadelta-betagammaneq=0$$
However, this leaves open the possibility that $T$ maps the upper half-plane to the $left$ half-plane.
This leaves you with two things to do. First, finish off the $cneq0$ case. (Hint: $Re T(i)>0$.) Second, do the (easier) $c=0$ case.
I feel that there must be an easier way of seeing this, but I've not been able to find one.
answered 21 hours ago
saulspatz
10.2k21323
1
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
add a comment |Â
up vote
1
down vote
I think I've got it. The transformation $T$ must take the boundary of the upper half-plane to the boundary of the right half-plane. That is, it must take the real axis to the imaginary axis. If $$T(z)=az+bover cz +d, ad-bcneq=0,tag1$$ we have that $xinmathbfR$ implies $$Refrac(ax+b)overline(cx+d)cx+d=0impliesRe(ax+b)overline(cx+d)=0$$
Now for given, $a,b,c,dinmathbfC, Re(ax+b)overline(cx+d)$ is a quadratic in $x$ that vanishes everywhere, so all the coefficients must be $0$. That is,
$$Re(aoverlinec)=Re(aoverlined+boverlinec)=Re(boverlined)=0tag2$$
Let us assume that $cneq0.$ Then we may divide numerator and denominator in $(1)$ by $c$, or what is the same thing, we may assume that $c=1,$ so $(2)$ becomes
$$Re(a)=Re(aoverlined+b)=Re(boverlined)=0tag3$$
From $c=1$ we have $T(-d)=infty,$ but $infty$ is on the imaginary axis so $dinmathbfR,$ and from $(3),$ we have $Re a=0$ and $Re(ad+b)=0,$ so that $Re b = 0.$ That is,
$$T(z) = ifracalpha z+betaz+d, text where alpha,beta,dinmathbfR, alpha d -betaneq0$$
which can obviously be re-written more symmetrically as
$$T(z) = ifracalpha z+betagamma z+ delta, text where alpha,beta,gamma deltainmathbfR, alphadelta-betagammaneq=0$$
However, this leaves open the possibility that $T$ maps the upper half-plane to the $left$ half-plane.
This leaves you with two things to do. First, finish off the $cneq0$ case. (Hint: $Re T(i)>0$.) Second, do the (easier) $c=0$ case.
I feel that there must be an easier way of seeing this, but I've not been able to find one.
answered 21 hours ago
saulspatz
10.2k21323
1
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I think I've got it. The transformation $T$ must take the boundary of the upper half-plane to the boundary of the right half-plane. That is, it must take the real axis to the imaginary axis. If $$T(z)=az+bover cz +d, ad-bcneq=0,tag1$$ we have that $xinmathbfR$ implies $$Refrac(ax+b)overline(cx+d)cx+d=0impliesRe(ax+b)overline(cx+d)=0$$
Now for given, $a,b,c,dinmathbfC, Re(ax+b)overline(cx+d)$ is a quadratic in $x$ that vanishes everywhere, so all the coefficients must be $0$. That is,
$$Re(aoverlinec)=Re(aoverlined+boverlinec)=Re(boverlined)=0tag2$$
Let us assume that $cneq0.$ Then we may divide numerator and denominator in $(1)$ by $c$, or what is the same thing, we may assume that $c=1,$ so $(2)$ becomes
$$Re(a)=Re(aoverlined+b)=Re(boverlined)=0tag3$$
From $c=1$ we have $T(-d)=infty,$ but $infty$ is on the imaginary axis so $dinmathbfR,$ and from $(3),$ we have $Re a=0$ and $Re(ad+b)=0,$ so that $Re b = 0.$ That is,
$$T(z) = ifracalpha z+betaz+d, text where alpha,beta,dinmathbfR, alpha d -betaneq0$$
which can obviously be re-written more symmetrically as
$$T(z) = ifracalpha z+betagamma z+ delta, text where alpha,beta,gamma deltainmathbfR, alphadelta-betagammaneq=0$$
However, this leaves open the possibility that $T$ maps the upper half-plane to the $left$ half-plane.
This leaves you with two things to do. First, finish off the $cneq0$ case. (Hint: $Re T(i)>0$.) Second, do the (easier) $c=0$ case.
I feel that there must be an easier way of seeing this, but I've not been able to find one.
answered 21 hours ago
saulspatz
10.2k21323
I think I've got it. The transformation $T$ must take the boundary of the upper half-plane to the boundary of the right half-plane. That is, it must take the real axis to the imaginary axis. If $$T(z)=az+bover cz +d, ad-bcneq=0,tag1$$ we have that $xinmathbfR$ implies $$Refrac(ax+b)overline(cx+d)cx+d=0impliesRe(ax+b)overline(cx+d)=0$$
Now for given, $a,b,c,dinmathbfC, Re(ax+b)overline(cx+d)$ is a quadratic in $x$ that vanishes everywhere, so all the coefficients must be $0$. That is,
$$Re(aoverlinec)=Re(aoverlined+boverlinec)=Re(boverlined)=0tag2$$
Let us assume that $cneq0.$ Then we may divide numerator and denominator in $(1)$ by $c$, or what is the same thing, we may assume that $c=1,$ so $(2)$ becomes
$$Re(a)=Re(aoverlined+b)=Re(boverlined)=0tag3$$
From $c=1$ we have $T(-d)=infty,$ but $infty$ is on the imaginary axis so $dinmathbfR,$ and from $(3),$ we have $Re a=0$ and $Re(ad+b)=0,$ so that $Re b = 0.$ That is,
$$T(z) = ifracalpha z+betaz+d, text where alpha,beta,dinmathbfR, alpha d -betaneq0$$
which can obviously be re-written more symmetrically as
$$T(z) = ifracalpha z+betagamma z+ delta, text where alpha,beta,gamma deltainmathbfR, alphadelta-betagammaneq=0$$
However, this leaves open the possibility that $T$ maps the upper half-plane to the $left$ half-plane.
This leaves you with two things to do. First, finish off the $cneq0$ case. (Hint: $Re T(i)>0$.) Second, do the (easier) $c=0$ case.
I feel that there must be an easier way of seeing this, but I've not been able to find one.
answered 21 hours ago
saulspatz
10.2k21323
answered 21 hours ago
saulspatz
10.2k21323
answered 21 hours ago
saulspatz
10.2k21323
answered 21 hours ago
saulspatz
10.2k21323
10.2k21323
1
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
add a comment |Â
1
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
1
1
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
Might be easier to start with the mapping of the upper half-plane to itself and then multiply by $-i$. From the formula for a linear-fractional transform mapping $z_1, z_2, z_3$ to $w_1, w_2, w_3$, we obtain that mappings of the real line to itself are given by $(a z + b)/(c z + d)$ with $a, b, c, d$ real. The condition that the direction is preserved is given by $T'(z) > 0$ for real $z$, or $a d - b c > 0$.
– Maxim
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
@Maxim T I don't have time to consider this closely just now, but on a quick read, it sounds goo. I'll think about it later. Thanks.
– saulspatz
17 hours ago
add a comment |Â
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6
How about you just rotate the plane?
– quid♦
yesterday
@quid Ok, Am I going to use this transformation $T(z)=e^itheta_0z$ then?.
– Isa
yesterday
Yes. And what is $theta_0$?
– David G. Stork
yesterday
@DavidG.Stork it's the angle of rotation, the principal argument
– Isa
yesterday
1
I'd be glad to tell you if I knew how. A Möbius transformation $T$ is of the form $$T(z) =az+bover cz+d$$ where $ad-bcneq 0,$ so the question apparently asks for conditions on $a,b,c,d.$ It looks easy if $c=0$ -- just a straightforward elaboration of the answers given in the comments -- but I haven't been able to either handle the $cneq0$ case or to prove that $c=0.$
– saulspatz
yesterday