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Image of the domain $|z|<1$, $0<textArg,z<pi/4$ under the mappimg $w=(2z+1)/(z+i)$
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I'm trying to find the image of the domain $D$ under the mapping $w=frac2z+1z+i$, where
$$D=leftz$$
I've tried two ways but in both I arrived to something very complex.
First way: Since $z=frac-iw+1w-2$, we have $left|frac-iw+1w-2right|<1$ and after some manipulation we arrive to $4u+2v<3$. Also we should have
$$0<arctan(frac-u^2-v^2+2u-vu-2v-2)<fracpi4$$
Now if $u-2v-2>0$ we have $$0<frac-u^2-v^2+2u-vu-2v-2<1$$ and if $u-2v-2<0$ we should have $-u^2-v^2+2u-v<0$
Second way: Let $w_1=z+i$. Then
$w_1=leftz-i$. If $w_2=frac1w_1$, then since $|z-i|<1$ we arrive to $2v+1<0$ and since $0<textArg(z-i)<fracpi4$ we arrive to $0<frac-v-u^2-v^2u<1$ so again we should consider two cases $u>0$ and $u<0$.
Could anyone help me to find the right image?
complex-analysis complex-numbers
edited Jul 29 at 10:43
rtybase
8,77221333
asked Jul 29 at 7:22
user547800
166210
add a comment |Â
up vote
1
down vote
favorite
I'm trying to find the image of the domain $D$ under the mapping $w=frac2z+1z+i$, where
$$D=leftz$$
I've tried two ways but in both I arrived to something very complex.
First way: Since $z=frac-iw+1w-2$, we have $left|frac-iw+1w-2right|<1$ and after some manipulation we arrive to $4u+2v<3$. Also we should have
$$0<arctan(frac-u^2-v^2+2u-vu-2v-2)<fracpi4$$
Now if $u-2v-2>0$ we have $$0<frac-u^2-v^2+2u-vu-2v-2<1$$ and if $u-2v-2<0$ we should have $-u^2-v^2+2u-v<0$
Second way: Let $w_1=z+i$. Then
$w_1=leftz-i$. If $w_2=frac1w_1$, then since $|z-i|<1$ we arrive to $2v+1<0$ and since $0<textArg(z-i)<fracpi4$ we arrive to $0<frac-v-u^2-v^2u<1$ so again we should consider two cases $u>0$ and $u<0$.
Could anyone help me to find the right image?
complex-analysis complex-numbers
edited Jul 29 at 10:43
rtybase
8,77221333
asked Jul 29 at 7:22
user547800
166210
Hint: Such mappings map lines and circles to lines and circles (but not necessarily lines to lines and circles to circles). Border of $D$ consists of two line segments and one circle arc ...
– Antoine
Jul 29 at 7:57
Using the fact that $$z=frac1-iww-2$$ the "first way" indeed allows to see that the condition $|z|<1$ is equivalent to $w=u+iv$ with $u$ and $v$ real and $$4u+2v<3$$ But the argument ccndition $$0<mathrmArgzeta<pi/4$$ reads $$0<Imzeta<Rezeta$$ hence, using the positive multiple of $z$ equal to $$zeta=(1-iw)(bar w-2)$$ one gets $$0<2u-u^2-v^2-v<u-2-2v$$ from where you can conclude, right?
– Did
Jul 29 at 8:14
@Antoine Two and one, or one and two?
– Did
Jul 29 at 8:17
@Did What do you mean by "using the positive multiple of $z$ equal to $zeta=(1-iw)(barw-2)$"?
– user547800
Jul 29 at 9:12
That $zeta$ is a positive multiple of $z$ (hence their arguments are the same).
– Did
Jul 29 at 9:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to find the image of the domain $D$ under the mapping $w=frac2z+1z+i$, where
$$D=leftz$$
I've tried two ways but in both I arrived to something very complex.
First way: Since $z=frac-iw+1w-2$, we have $left|frac-iw+1w-2right|<1$ and after some manipulation we arrive to $4u+2v<3$. Also we should have
$$0<arctan(frac-u^2-v^2+2u-vu-2v-2)<fracpi4$$
Now if $u-2v-2>0$ we have $$0<frac-u^2-v^2+2u-vu-2v-2<1$$ and if $u-2v-2<0$ we should have $-u^2-v^2+2u-v<0$
Second way: Let $w_1=z+i$. Then
$w_1=leftz-i$. If $w_2=frac1w_1$, then since $|z-i|<1$ we arrive to $2v+1<0$ and since $0<textArg(z-i)<fracpi4$ we arrive to $0<frac-v-u^2-v^2u<1$ so again we should consider two cases $u>0$ and $u<0$.
Could anyone help me to find the right image?
complex-analysis complex-numbers
edited Jul 29 at 10:43
rtybase
8,77221333
asked Jul 29 at 7:22
user547800
166210
I'm trying to find the image of the domain $D$ under the mapping $w=frac2z+1z+i$, where
$$D=leftz$$
I've tried two ways but in both I arrived to something very complex.
First way: Since $z=frac-iw+1w-2$, we have $left|frac-iw+1w-2right|<1$ and after some manipulation we arrive to $4u+2v<3$. Also we should have
$$0<arctan(frac-u^2-v^2+2u-vu-2v-2)<fracpi4$$
Now if $u-2v-2>0$ we have $$0<frac-u^2-v^2+2u-vu-2v-2<1$$ and if $u-2v-2<0$ we should have $-u^2-v^2+2u-v<0$
Second way: Let $w_1=z+i$. Then
$w_1=leftz-i$. If $w_2=frac1w_1$, then since $|z-i|<1$ we arrive to $2v+1<0$ and since $0<textArg(z-i)<fracpi4$ we arrive to $0<frac-v-u^2-v^2u<1$ so again we should consider two cases $u>0$ and $u<0$.
Could anyone help me to find the right image?
complex-analysis complex-numbers
edited Jul 29 at 10:43
rtybase
8,77221333
asked Jul 29 at 7:22
user547800
166210
edited Jul 29 at 10:43
rtybase
8,77221333
edited Jul 29 at 10:43
rtybase
8,77221333
edited Jul 29 at 10:43
rtybase
8,77221333
8,77221333
asked Jul 29 at 7:22
user547800
166210
asked Jul 29 at 7:22
user547800
166210
asked Jul 29 at 7:22
user547800
166210
166210
Hint: Such mappings map lines and circles to lines and circles (but not necessarily lines to lines and circles to circles). Border of $D$ consists of two line segments and one circle arc ...
– Antoine
Jul 29 at 7:57
Using the fact that $$z=frac1-iww-2$$ the "first way" indeed allows to see that the condition $|z|<1$ is equivalent to $w=u+iv$ with $u$ and $v$ real and $$4u+2v<3$$ But the argument ccndition $$0<mathrmArgzeta<pi/4$$ reads $$0<Imzeta<Rezeta$$ hence, using the positive multiple of $z$ equal to $$zeta=(1-iw)(bar w-2)$$ one gets $$0<2u-u^2-v^2-v<u-2-2v$$ from where you can conclude, right?
– Did
Jul 29 at 8:14
@Antoine Two and one, or one and two?
– Did
Jul 29 at 8:17
@Did What do you mean by "using the positive multiple of $z$ equal to $zeta=(1-iw)(barw-2)$"?
– user547800
Jul 29 at 9:12
That $zeta$ is a positive multiple of $z$ (hence their arguments are the same).
– Did
Jul 29 at 9:43
add a comment |Â
Hint: Such mappings map lines and circles to lines and circles (but not necessarily lines to lines and circles to circles). Border of $D$ consists of two line segments and one circle arc ...
– Antoine
Jul 29 at 7:57
Using the fact that $$z=frac1-iww-2$$ the "first way" indeed allows to see that the condition $|z|<1$ is equivalent to $w=u+iv$ with $u$ and $v$ real and $$4u+2v<3$$ But the argument ccndition $$0<mathrmArgzeta<pi/4$$ reads $$0<Imzeta<Rezeta$$ hence, using the positive multiple of $z$ equal to $$zeta=(1-iw)(bar w-2)$$ one gets $$0<2u-u^2-v^2-v<u-2-2v$$ from where you can conclude, right?
– Did
Jul 29 at 8:14
@Antoine Two and one, or one and two?
– Did
Jul 29 at 8:17
@Did What do you mean by "using the positive multiple of $z$ equal to $zeta=(1-iw)(barw-2)$"?
– user547800
Jul 29 at 9:12
That $zeta$ is a positive multiple of $z$ (hence their arguments are the same).
– Did
Jul 29 at 9:43
Hint: Such mappings map lines and circles to lines and circles (but not necessarily lines to lines and circles to circles). Border of $D$ consists of two line segments and one circle arc ...
– Antoine
Jul 29 at 7:57
Hint: Such mappings map lines and circles to lines and circles (but not necessarily lines to lines and circles to circles). Border of $D$ consists of two line segments and one circle arc ...
– Antoine
Jul 29 at 7:57
Using the fact that $$z=frac1-iww-2$$ the "first way" indeed allows to see that the condition $|z|<1$ is equivalent to $w=u+iv$ with $u$ and $v$ real and $$4u+2v<3$$ But the argument ccndition $$0<mathrmArgzeta<pi/4$$ reads $$0<Imzeta<Rezeta$$ hence, using the positive multiple of $z$ equal to $$zeta=(1-iw)(bar w-2)$$ one gets $$0<2u-u^2-v^2-v<u-2-2v$$ from where you can conclude, right?
– Did
Jul 29 at 8:14
Using the fact that $$z=frac1-iww-2$$ the "first way" indeed allows to see that the condition $|z|<1$ is equivalent to $w=u+iv$ with $u$ and $v$ real and $$4u+2v<3$$ But the argument ccndition $$0<mathrmArgzeta<pi/4$$ reads $$0<Imzeta<Rezeta$$ hence, using the positive multiple of $z$ equal to $$zeta=(1-iw)(bar w-2)$$ one gets $$0<2u-u^2-v^2-v<u-2-2v$$ from where you can conclude, right?
– Did
Jul 29 at 8:14
@Antoine Two and one, or one and two?
– Did
Jul 29 at 8:17
@Antoine Two and one, or one and two?
– Did
Jul 29 at 8:17
@Did What do you mean by "using the positive multiple of $z$ equal to $zeta=(1-iw)(barw-2)$"?
– user547800
Jul 29 at 9:12
@Did What do you mean by "using the positive multiple of $z$ equal to $zeta=(1-iw)(barw-2)$"?
– user547800
Jul 29 at 9:12
That $zeta$ is a positive multiple of $z$ (hence their arguments are the same).
– Did
Jul 29 at 9:43
That $zeta$ is a positive multiple of $z$ (hence their arguments are the same).
– Did
Jul 29 at 9:43
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
-1
down vote
We map $D$ to unit disk $U=<1$ with $w_2=z^8$ and $w_1=frac2z+1z+i$ maps $U$ to half-plane with border $y=-2x+frac32$, it is
$$zinmathbbC:y<-2x+frac32$$
then you want the map $w_2ow_1$.
answered Jul 29 at 8:23
Lolita
52318
Why $z^8$.....?
– user547800
Jul 29 at 9:18
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
add a comment |Â
up vote
-1
down vote
Your solution is correct, the area lies between two circles
$$-u^2-v^2+2u-v<0~~~~~;~~~~~-u^2-v^2+2u-v<u-2v-2$$
which you found by restrictions, they are
$$|z-(1-fraci2)|>fracsqrt52$$
$$|z-(frac12-fraci2)|<frac32$$
this area lies between two rays $u>0$ and $2u+v<frac32$.
answered Jul 29 at 8:50
Lolita
52318
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
We map $D$ to unit disk $U=<1$ with $w_2=z^8$ and $w_1=frac2z+1z+i$ maps $U$ to half-plane with border $y=-2x+frac32$, it is
$$zinmathbbC:y<-2x+frac32$$
then you want the map $w_2ow_1$.
answered Jul 29 at 8:23
Lolita
52318
Why $z^8$.....?
– user547800
Jul 29 at 9:18
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
add a comment |Â
up vote
-1
down vote
We map $D$ to unit disk $U=<1$ with $w_2=z^8$ and $w_1=frac2z+1z+i$ maps $U$ to half-plane with border $y=-2x+frac32$, it is
$$zinmathbbC:y<-2x+frac32$$
then you want the map $w_2ow_1$.
answered Jul 29 at 8:23
Lolita
52318
Why $z^8$.....?
– user547800
Jul 29 at 9:18
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
We map $D$ to unit disk $U=<1$ with $w_2=z^8$ and $w_1=frac2z+1z+i$ maps $U$ to half-plane with border $y=-2x+frac32$, it is
$$zinmathbbC:y<-2x+frac32$$
then you want the map $w_2ow_1$.
answered Jul 29 at 8:23
Lolita
52318
We map $D$ to unit disk $U=<1$ with $w_2=z^8$ and $w_1=frac2z+1z+i$ maps $U$ to half-plane with border $y=-2x+frac32$, it is
$$zinmathbbC:y<-2x+frac32$$
then you want the map $w_2ow_1$.
answered Jul 29 at 8:23
Lolita
52318
answered Jul 29 at 8:23
Lolita
52318
answered Jul 29 at 8:23
Lolita
52318
answered Jul 29 at 8:23
Lolita
52318
52318
Why $z^8$.....?
– user547800
Jul 29 at 9:18
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
add a comment |Â
Why $z^8$.....?
– user547800
Jul 29 at 9:18
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
Why $z^8$.....?
– user547800
Jul 29 at 9:18
Why $z^8$.....?
– user547800
Jul 29 at 9:18
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
this maps angle 45^o to 2pi
– Lolita
Jul 29 at 9:31
add a comment |Â
up vote
-1
down vote
Your solution is correct, the area lies between two circles
$$-u^2-v^2+2u-v<0~~~~~;~~~~~-u^2-v^2+2u-v<u-2v-2$$
which you found by restrictions, they are
$$|z-(1-fraci2)|>fracsqrt52$$
$$|z-(frac12-fraci2)|<frac32$$
this area lies between two rays $u>0$ and $2u+v<frac32$.
answered Jul 29 at 8:50
Lolita
52318
add a comment |Â
up vote
-1
down vote
Your solution is correct, the area lies between two circles
$$-u^2-v^2+2u-v<0~~~~~;~~~~~-u^2-v^2+2u-v<u-2v-2$$
which you found by restrictions, they are
$$|z-(1-fraci2)|>fracsqrt52$$
$$|z-(frac12-fraci2)|<frac32$$
this area lies between two rays $u>0$ and $2u+v<frac32$.
answered Jul 29 at 8:50
Lolita
52318
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Your solution is correct, the area lies between two circles
$$-u^2-v^2+2u-v<0~~~~~;~~~~~-u^2-v^2+2u-v<u-2v-2$$
which you found by restrictions, they are
$$|z-(1-fraci2)|>fracsqrt52$$
$$|z-(frac12-fraci2)|<frac32$$
this area lies between two rays $u>0$ and $2u+v<frac32$.
answered Jul 29 at 8:50
Lolita
52318
Your solution is correct, the area lies between two circles
$$-u^2-v^2+2u-v<0~~~~~;~~~~~-u^2-v^2+2u-v<u-2v-2$$
which you found by restrictions, they are
$$|z-(1-fraci2)|>fracsqrt52$$
$$|z-(frac12-fraci2)|<frac32$$
this area lies between two rays $u>0$ and $2u+v<frac32$.
answered Jul 29 at 8:50
Lolita
52318
edited Jul 29 at 8:55
answered Jul 29 at 8:50
Lolita
52318
answered Jul 29 at 8:50
Lolita
52318
answered Jul 29 at 8:50
Lolita
52318
52318
add a comment |Â
add a comment |Â
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Hint: Such mappings map lines and circles to lines and circles (but not necessarily lines to lines and circles to circles). Border of $D$ consists of two line segments and one circle arc ...
– Antoine
Jul 29 at 7:57
Using the fact that $$z=frac1-iww-2$$ the "first way" indeed allows to see that the condition $|z|<1$ is equivalent to $w=u+iv$ with $u$ and $v$ real and $$4u+2v<3$$ But the argument ccndition $$0<mathrmArgzeta<pi/4$$ reads $$0<Imzeta<Rezeta$$ hence, using the positive multiple of $z$ equal to $$zeta=(1-iw)(bar w-2)$$ one gets $$0<2u-u^2-v^2-v<u-2-2v$$ from where you can conclude, right?
– Did
Jul 29 at 8:14
@Antoine Two and one, or one and two?
– Did
Jul 29 at 8:17
@Did What do you mean by "using the positive multiple of $z$ equal to $zeta=(1-iw)(barw-2)$"?
– user547800
Jul 29 at 9:12
That $zeta$ is a positive multiple of $z$ (hence their arguments are the same).
– Did
Jul 29 at 9:43