Mixing
Show that the set $z$ satisfying $|z-z_0|=rho|z-z_1|$ ($rho neq 1$) is a circle.
$begingroup$
Prove that for $rho ge 0$, $rho neq 1$ and fix $z_0,z_1 in Bbb C$. Show that the set $z in Bbb C$ satisfying $|z-z_0|=rho|z-z_1|$ is a circle.
From the given condition I have reached that $(1-rho^2)|z|^2 -2mathfrak{Re} (bar z (z_0-rho^2 z_1)) + |z_0|^2 -rho^2|z_1|^2=0$
But I do not get my desired result.
complex-analysis complex-numbers
edited Aug 6 '15 at 11:40
mrf
37.5k54685
asked Aug 5 '15 at 19:27
user152715user152715
1
$endgroup$
|
show 2 more comments
$begingroup$
Prove that for $rho ge 0$, $rho neq 1$ and fix $z_0,z_1 in Bbb C$. Show that the set $z in Bbb C$ satisfying $|z-z_0|=rho|z-z_1|$ is a circle.
From the given condition I have reached that $(1-rho^2)|z|^2 -2mathfrak{Re} (bar z (z_0-rho^2 z_1)) + |z_0|^2 -rho^2|z_1|^2=0$
But I do not get my desired result.
complex-analysis complex-numbers
edited Aug 6 '15 at 11:40
mrf
37.5k54685
asked Aug 5 '15 at 19:27
user152715user152715
1
$endgroup$
$begingroup$
Why don't you tackle it geometrically? Draw a line from $z_0$ to $z_1$, draw the points on this line satisfying the equation (there must be two), then find the centre point between these two, the equation of that circle, and show all points satisfy the equation?
$endgroup$
– Colm Bhandal
Aug 5 '15 at 19:41
$begingroup$
Then I have to show that no other point satisfy that eqn. Will you do it elaborately and give me the answer? Because I think doing algebraic way will be shorter.
$endgroup$
– user152715
Aug 5 '15 at 19:44
$begingroup$
Have you already treated Möbius transformations?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:50
$begingroup$
No, I have not used it. This is first chapter problem in Gamelin's book and I want to do it in an elementary way.
$endgroup$
– user152715
Aug 5 '15 at 19:52
1
$begingroup$
Okay, then do it per pedes. So you found that the condition is equivalent to $alvert zrvert^2 - 2 operatorname{Re}(boverline{z}) + c = 0$ for some real $a,c$ with $a neq 0$ and $b in mathbb{C}$. When is that sort of equation the defining equation for a circle?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:55
|
show 2 more comments
$begingroup$
Prove that for $rho ge 0$, $rho neq 1$ and fix $z_0,z_1 in Bbb C$. Show that the set $z in Bbb C$ satisfying $|z-z_0|=rho|z-z_1|$ is a circle.
From the given condition I have reached that $(1-rho^2)|z|^2 -2mathfrak{Re} (bar z (z_0-rho^2 z_1)) + |z_0|^2 -rho^2|z_1|^2=0$
But I do not get my desired result.
complex-analysis complex-numbers
edited Aug 6 '15 at 11:40
mrf
37.5k54685
asked Aug 5 '15 at 19:27
user152715user152715
1
$endgroup$
Prove that for $rho ge 0$, $rho neq 1$ and fix $z_0,z_1 in Bbb C$. Show that the set $z in Bbb C$ satisfying $|z-z_0|=rho|z-z_1|$ is a circle.
From the given condition I have reached that $(1-rho^2)|z|^2 -2mathfrak{Re} (bar z (z_0-rho^2 z_1)) + |z_0|^2 -rho^2|z_1|^2=0$
But I do not get my desired result.
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Aug 6 '15 at 11:40
mrf
37.5k54685
asked Aug 5 '15 at 19:27
user152715user152715
1
edited Aug 6 '15 at 11:40
mrf
37.5k54685
asked Aug 5 '15 at 19:27
user152715user152715
1
edited Aug 6 '15 at 11:40
mrf
37.5k54685
edited Aug 6 '15 at 11:40
mrf
37.5k54685
edited Aug 6 '15 at 11:40
mrf
37.5k54685
37.5k54685
asked Aug 5 '15 at 19:27
user152715user152715
1
asked Aug 5 '15 at 19:27
user152715user152715
1
asked Aug 5 '15 at 19:27
user152715user152715
1
1
$begingroup$
Why don't you tackle it geometrically? Draw a line from $z_0$ to $z_1$, draw the points on this line satisfying the equation (there must be two), then find the centre point between these two, the equation of that circle, and show all points satisfy the equation?
$endgroup$
– Colm Bhandal
Aug 5 '15 at 19:41
$begingroup$
Then I have to show that no other point satisfy that eqn. Will you do it elaborately and give me the answer? Because I think doing algebraic way will be shorter.
$endgroup$
– user152715
Aug 5 '15 at 19:44
$begingroup$
Have you already treated Möbius transformations?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:50
$begingroup$
No, I have not used it. This is first chapter problem in Gamelin's book and I want to do it in an elementary way.
$endgroup$
– user152715
Aug 5 '15 at 19:52
1
$begingroup$
Okay, then do it per pedes. So you found that the condition is equivalent to $alvert zrvert^2 - 2 operatorname{Re}(boverline{z}) + c = 0$ for some real $a,c$ with $a neq 0$ and $b in mathbb{C}$. When is that sort of equation the defining equation for a circle?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:55
|
show 2 more comments
$begingroup$
Why don't you tackle it geometrically? Draw a line from $z_0$ to $z_1$, draw the points on this line satisfying the equation (there must be two), then find the centre point between these two, the equation of that circle, and show all points satisfy the equation?
$endgroup$
– Colm Bhandal
Aug 5 '15 at 19:41
$begingroup$
Then I have to show that no other point satisfy that eqn. Will you do it elaborately and give me the answer? Because I think doing algebraic way will be shorter.
$endgroup$
– user152715
Aug 5 '15 at 19:44
$begingroup$
Have you already treated Möbius transformations?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:50
$begingroup$
No, I have not used it. This is first chapter problem in Gamelin's book and I want to do it in an elementary way.
$endgroup$
– user152715
Aug 5 '15 at 19:52
1
$begingroup$
Okay, then do it per pedes. So you found that the condition is equivalent to $alvert zrvert^2 - 2 operatorname{Re}(boverline{z}) + c = 0$ for some real $a,c$ with $a neq 0$ and $b in mathbb{C}$. When is that sort of equation the defining equation for a circle?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:55
$begingroup$
Why don't you tackle it geometrically? Draw a line from $z_0$ to $z_1$, draw the points on this line satisfying the equation (there must be two), then find the centre point between these two, the equation of that circle, and show all points satisfy the equation?
$endgroup$
– Colm Bhandal
Aug 5 '15 at 19:41
$begingroup$
Why don't you tackle it geometrically? Draw a line from $z_0$ to $z_1$, draw the points on this line satisfying the equation (there must be two), then find the centre point between these two, the equation of that circle, and show all points satisfy the equation?
$endgroup$
– Colm Bhandal
Aug 5 '15 at 19:41
$begingroup$
Then I have to show that no other point satisfy that eqn. Will you do it elaborately and give me the answer? Because I think doing algebraic way will be shorter.
$endgroup$
– user152715
Aug 5 '15 at 19:44
$begingroup$
Then I have to show that no other point satisfy that eqn. Will you do it elaborately and give me the answer? Because I think doing algebraic way will be shorter.
$endgroup$
– user152715
Aug 5 '15 at 19:44
$begingroup$
Have you already treated Möbius transformations?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:50
$begingroup$
Have you already treated Möbius transformations?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:50
$begingroup$
No, I have not used it. This is first chapter problem in Gamelin's book and I want to do it in an elementary way.
$endgroup$
– user152715
Aug 5 '15 at 19:52
$begingroup$
No, I have not used it. This is first chapter problem in Gamelin's book and I want to do it in an elementary way.
$endgroup$
– user152715
Aug 5 '15 at 19:52
1
1
$begingroup$
Okay, then do it per pedes. So you found that the condition is equivalent to $alvert zrvert^2 - 2 operatorname{Re}(boverline{z}) + c = 0$ for some real $a,c$ with $a neq 0$ and $b in mathbb{C}$. When is that sort of equation the defining equation for a circle?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:55
$begingroup$
Okay, then do it per pedes. So you found that the condition is equivalent to $alvert zrvert^2 - 2 operatorname{Re}(boverline{z}) + c = 0$ for some real $a,c$ with $a neq 0$ and $b in mathbb{C}$. When is that sort of equation the defining equation for a circle?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:55
|
show 2 more comments
4 Answers
4
active
oldest
votes
$begingroup$
From where you left off
$$(1-rho^{2})|z|^{2}-lbrace z^{*}(z_{0}-rho^{2}z_{1})+z(z_{0}-rho^{2}z_{1})^{*}rbrace=rho^{2}|z_{1}|^{2}-|z_{0}|^{2}$$
Dividing by $(1-rho^{2})$ and completing the square
$$|z-frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2}=frac{rho^{2}|z_{1}|^{2}-|z_{0}|^{2}}{1-rho^{2}}+|frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2} $$
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
$endgroup$
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
add a comment |
$begingroup$
WOLOG, we may assume $z_1 =0$. Then we are looking at $|z-z_0|=rho |z|$. Next, by rotation, we may assume that $a=z_0$ is real and positive.
Now, let $z=x+iy$ and square the equation to get
$(x-a)^2 +y^2=rho^2 (x^2 +y^2)$. When this is expanded, a conic section is obtained with equal coefficients for $x$ and $y$. This is a circle.
Alternatively, and using what you wrote, let $z=x+iy$ and find that the coefficients of $x^2$ and $y^2$ are the same, and there is $xy$ term. This gives a circle.
answered Aug 5 '15 at 19:59
polymath257polymath257
141
$endgroup$
add a comment |
$begingroup$
Hint..you can write $z=x+iy$, $z_0=a+ib$, $z_1=c+id$ for real $x,y,a,b,c,d$ and apply your condition so that you get$$(x-a)^2+(y-b)^2=rho^2[(x-c)^2+(y-d)^2]$$ which will lead to the equation of a circle
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
$endgroup$
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
$begingroup$
I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
$begingroup$
@ColmBhandal: yes! Thanks
$endgroup$
– David Quinn
Aug 5 '15 at 20:06
$begingroup$
@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
$endgroup$
– David Quinn
Aug 5 '15 at 20:12
$begingroup$
@DavidQuinn which are the radius and the center of the circle?
$endgroup$
– Maggie94
Jan 15 at 14:48
add a comment |
$begingroup$
Begin by noting that $langle x+iy,a+ib rangle = xa+yb = text{Re}left((x+iy)(a-ib)right)$ so if we think of $z$ and $w$ as two-dimensional vectors then the complex algebra permits a natural formulation of the Euclidean inner-product in the plane:
$$ langle v,w rangle = text{Re} (v bar{w}) $$
On the other hand, we have $|z|^2 = z bar{z}$ and the many wonderful properties of the complex conjugate. Suppose $|z-z_o| = rho|z-z_1|$ for some $rho>0$ (if I say $rho>0$ then it must be real !). Square the given condition and use properties of complex conjugation:
$$ (z-z_o)(bar{z}-bar{z_o}) = rho^2(z-z_1)(bar{z}-bar{z_1})$$
or
$$ zbar{z}-zbar{z_o}-z_obar{z}+z_obar{z_o} = rho^2left( zbar{z}-zbar{z_1}-z_1bar{z}+z_1bar{z_1}right)$$
which gives
$$ |z|^2-zbar{z_o}-z_obar{z}+|z_o|^2 = rho^2left( |z|^2-zbar{z_1}-z_1bar{z}+|z_1|^2right)$$
which we can rearrange to
$$ rho^2zbar{z_1}+rho^2z_1bar{z}-zbar{z_o}-z_obar{z} =
(rho^2-1)|z|^2 +rho^2|z_1|^2- |z_o|^2$$
Ok, I'm tired of $rho$, I'm setting $rho=1$ in which case:
$$ zbar{z_1}+z_1bar{z}-zbar{z_o}-z_obar{z} =
|z_1|^2- |z_o|^2$$
which gives, using $text{Re}(z) = frac{1}{2}(z + bar{z})$,
$$ z(overline{z_1-z_o})+bar{z}(z_1-z_o) = 2text{Re}left(z(overline{z_1-z_o})right) = |z_1|^2- |z_o|^2$$
or,
$$ langle z, z_1-z_o rangle = frac{1}{2}left(|z_1|^2- |z_o|^2right) $$
Now, if we have $|z_o| = |z_1|$ then it follows:
$$ langle z, z_1-z_o rangle = 0 $$
which says $z$ points in a direction perpendicular to the line-segment from $z_0$ to $z_1$. In any event, you can use the things I've done here to find the answer.
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
$endgroup$
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From where you left off
$$(1-rho^{2})|z|^{2}-lbrace z^{*}(z_{0}-rho^{2}z_{1})+z(z_{0}-rho^{2}z_{1})^{*}rbrace=rho^{2}|z_{1}|^{2}-|z_{0}|^{2}$$
Dividing by $(1-rho^{2})$ and completing the square
$$|z-frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2}=frac{rho^{2}|z_{1}|^{2}-|z_{0}|^{2}}{1-rho^{2}}+|frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2} $$
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
$endgroup$
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
add a comment |
$begingroup$
From where you left off
$$(1-rho^{2})|z|^{2}-lbrace z^{*}(z_{0}-rho^{2}z_{1})+z(z_{0}-rho^{2}z_{1})^{*}rbrace=rho^{2}|z_{1}|^{2}-|z_{0}|^{2}$$
Dividing by $(1-rho^{2})$ and completing the square
$$|z-frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2}=frac{rho^{2}|z_{1}|^{2}-|z_{0}|^{2}}{1-rho^{2}}+|frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2} $$
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
$endgroup$
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
add a comment |
$begingroup$
From where you left off
$$(1-rho^{2})|z|^{2}-lbrace z^{*}(z_{0}-rho^{2}z_{1})+z(z_{0}-rho^{2}z_{1})^{*}rbrace=rho^{2}|z_{1}|^{2}-|z_{0}|^{2}$$
Dividing by $(1-rho^{2})$ and completing the square
$$|z-frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2}=frac{rho^{2}|z_{1}|^{2}-|z_{0}|^{2}}{1-rho^{2}}+|frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2} $$
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
$endgroup$
From where you left off
$$(1-rho^{2})|z|^{2}-lbrace z^{*}(z_{0}-rho^{2}z_{1})+z(z_{0}-rho^{2}z_{1})^{*}rbrace=rho^{2}|z_{1}|^{2}-|z_{0}|^{2}$$
Dividing by $(1-rho^{2})$ and completing the square
$$|z-frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2}=frac{rho^{2}|z_{1}|^{2}-|z_{0}|^{2}}{1-rho^{2}}+|frac{1}{(1-rho^{2})}(z_{0}-rho^{2}z_{1})|^{2} $$
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
answered Aug 5 '15 at 20:15
TuckerTucker
1,510311
1,510311
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
add a comment |
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
we have a winner.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:19
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
$begingroup$
And you have to show that the RHS is non negative and it is true if $z_0 neq z_1$
$endgroup$
– user152715
Aug 5 '15 at 20:25
add a comment |
$begingroup$
WOLOG, we may assume $z_1 =0$. Then we are looking at $|z-z_0|=rho |z|$. Next, by rotation, we may assume that $a=z_0$ is real and positive.
Now, let $z=x+iy$ and square the equation to get
$(x-a)^2 +y^2=rho^2 (x^2 +y^2)$. When this is expanded, a conic section is obtained with equal coefficients for $x$ and $y$. This is a circle.
Alternatively, and using what you wrote, let $z=x+iy$ and find that the coefficients of $x^2$ and $y^2$ are the same, and there is $xy$ term. This gives a circle.
answered Aug 5 '15 at 19:59
polymath257polymath257
141
$endgroup$
add a comment |
$begingroup$
WOLOG, we may assume $z_1 =0$. Then we are looking at $|z-z_0|=rho |z|$. Next, by rotation, we may assume that $a=z_0$ is real and positive.
Now, let $z=x+iy$ and square the equation to get
$(x-a)^2 +y^2=rho^2 (x^2 +y^2)$. When this is expanded, a conic section is obtained with equal coefficients for $x$ and $y$. This is a circle.
Alternatively, and using what you wrote, let $z=x+iy$ and find that the coefficients of $x^2$ and $y^2$ are the same, and there is $xy$ term. This gives a circle.
answered Aug 5 '15 at 19:59
polymath257polymath257
141
$endgroup$
add a comment |
$begingroup$
WOLOG, we may assume $z_1 =0$. Then we are looking at $|z-z_0|=rho |z|$. Next, by rotation, we may assume that $a=z_0$ is real and positive.
Now, let $z=x+iy$ and square the equation to get
$(x-a)^2 +y^2=rho^2 (x^2 +y^2)$. When this is expanded, a conic section is obtained with equal coefficients for $x$ and $y$. This is a circle.
Alternatively, and using what you wrote, let $z=x+iy$ and find that the coefficients of $x^2$ and $y^2$ are the same, and there is $xy$ term. This gives a circle.
answered Aug 5 '15 at 19:59
polymath257polymath257
141
$endgroup$
WOLOG, we may assume $z_1 =0$. Then we are looking at $|z-z_0|=rho |z|$. Next, by rotation, we may assume that $a=z_0$ is real and positive.
Now, let $z=x+iy$ and square the equation to get
$(x-a)^2 +y^2=rho^2 (x^2 +y^2)$. When this is expanded, a conic section is obtained with equal coefficients for $x$ and $y$. This is a circle.
Alternatively, and using what you wrote, let $z=x+iy$ and find that the coefficients of $x^2$ and $y^2$ are the same, and there is $xy$ term. This gives a circle.
answered Aug 5 '15 at 19:59
polymath257polymath257
141
answered Aug 5 '15 at 19:59
polymath257polymath257
141
answered Aug 5 '15 at 19:59
polymath257polymath257
141
answered Aug 5 '15 at 19:59
polymath257polymath257
141
141
add a comment |
add a comment |
$begingroup$
Hint..you can write $z=x+iy$, $z_0=a+ib$, $z_1=c+id$ for real $x,y,a,b,c,d$ and apply your condition so that you get$$(x-a)^2+(y-b)^2=rho^2[(x-c)^2+(y-d)^2]$$ which will lead to the equation of a circle
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
$endgroup$
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
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I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
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@ColmBhandal: yes! Thanks
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– David Quinn
Aug 5 '15 at 20:06
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@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
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– David Quinn
Aug 5 '15 at 20:12
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@DavidQuinn which are the radius and the center of the circle?
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– Maggie94
Jan 15 at 14:48
add a comment |
$begingroup$
Hint..you can write $z=x+iy$, $z_0=a+ib$, $z_1=c+id$ for real $x,y,a,b,c,d$ and apply your condition so that you get$$(x-a)^2+(y-b)^2=rho^2[(x-c)^2+(y-d)^2]$$ which will lead to the equation of a circle
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
$endgroup$
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
$begingroup$
I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
$begingroup$
@ColmBhandal: yes! Thanks
$endgroup$
– David Quinn
Aug 5 '15 at 20:06
$begingroup$
@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
$endgroup$
– David Quinn
Aug 5 '15 at 20:12
$begingroup$
@DavidQuinn which are the radius and the center of the circle?
$endgroup$
– Maggie94
Jan 15 at 14:48
add a comment |
$begingroup$
Hint..you can write $z=x+iy$, $z_0=a+ib$, $z_1=c+id$ for real $x,y,a,b,c,d$ and apply your condition so that you get$$(x-a)^2+(y-b)^2=rho^2[(x-c)^2+(y-d)^2]$$ which will lead to the equation of a circle
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
$endgroup$
Hint..you can write $z=x+iy$, $z_0=a+ib$, $z_1=c+id$ for real $x,y,a,b,c,d$ and apply your condition so that you get$$(x-a)^2+(y-b)^2=rho^2[(x-c)^2+(y-d)^2]$$ which will lead to the equation of a circle
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
edited Aug 5 '15 at 20:06
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
answered Aug 5 '15 at 19:58
David QuinnDavid Quinn
24k21141
24k21141
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
$begingroup$
I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
$begingroup$
@ColmBhandal: yes! Thanks
$endgroup$
– David Quinn
Aug 5 '15 at 20:06
$begingroup$
@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
$endgroup$
– David Quinn
Aug 5 '15 at 20:12
$begingroup$
@DavidQuinn which are the radius and the center of the circle?
$endgroup$
– Maggie94
Jan 15 at 14:48
add a comment |
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
$begingroup$
I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
$begingroup$
@ColmBhandal: yes! Thanks
$endgroup$
– David Quinn
Aug 5 '15 at 20:06
$begingroup$
@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
$endgroup$
– David Quinn
Aug 5 '15 at 20:12
$begingroup$
@DavidQuinn which are the radius and the center of the circle?
$endgroup$
– Maggie94
Jan 15 at 14:48
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
$begingroup$
Is this a hint?? When I have done that calculation then there is no point saying me to put $z,z_0$ etc etc.
$endgroup$
– user152715
Aug 5 '15 at 20:02
$begingroup$
I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
$begingroup$
I think there should be an $x - c$ on the RHS instead of $-c$. Otherwise this seems like the most elegant, concise solution.
$endgroup$
– Colm Bhandal
Aug 5 '15 at 20:05
$begingroup$
@ColmBhandal: yes! Thanks
$endgroup$
– David Quinn
Aug 5 '15 at 20:06
$begingroup$
@ColmBhandal: yes! Thanks
$endgroup$
– David Quinn
Aug 5 '15 at 20:06
$begingroup$
@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
$endgroup$
– David Quinn
Aug 5 '15 at 20:12
$begingroup$
@user152715 it is only intended as a hint in the sense that this will lead to the solution but does not constitute the full solution
$endgroup$
– David Quinn
Aug 5 '15 at 20:12
$begingroup$
@DavidQuinn which are the radius and the center of the circle?
$endgroup$
– Maggie94
Jan 15 at 14:48
$begingroup$
@DavidQuinn which are the radius and the center of the circle?
$endgroup$
– Maggie94
Jan 15 at 14:48
add a comment |
$begingroup$
Begin by noting that $langle x+iy,a+ib rangle = xa+yb = text{Re}left((x+iy)(a-ib)right)$ so if we think of $z$ and $w$ as two-dimensional vectors then the complex algebra permits a natural formulation of the Euclidean inner-product in the plane:
$$ langle v,w rangle = text{Re} (v bar{w}) $$
On the other hand, we have $|z|^2 = z bar{z}$ and the many wonderful properties of the complex conjugate. Suppose $|z-z_o| = rho|z-z_1|$ for some $rho>0$ (if I say $rho>0$ then it must be real !). Square the given condition and use properties of complex conjugation:
$$ (z-z_o)(bar{z}-bar{z_o}) = rho^2(z-z_1)(bar{z}-bar{z_1})$$
or
$$ zbar{z}-zbar{z_o}-z_obar{z}+z_obar{z_o} = rho^2left( zbar{z}-zbar{z_1}-z_1bar{z}+z_1bar{z_1}right)$$
which gives
$$ |z|^2-zbar{z_o}-z_obar{z}+|z_o|^2 = rho^2left( |z|^2-zbar{z_1}-z_1bar{z}+|z_1|^2right)$$
which we can rearrange to
$$ rho^2zbar{z_1}+rho^2z_1bar{z}-zbar{z_o}-z_obar{z} =
(rho^2-1)|z|^2 +rho^2|z_1|^2- |z_o|^2$$
Ok, I'm tired of $rho$, I'm setting $rho=1$ in which case:
$$ zbar{z_1}+z_1bar{z}-zbar{z_o}-z_obar{z} =
|z_1|^2- |z_o|^2$$
which gives, using $text{Re}(z) = frac{1}{2}(z + bar{z})$,
$$ z(overline{z_1-z_o})+bar{z}(z_1-z_o) = 2text{Re}left(z(overline{z_1-z_o})right) = |z_1|^2- |z_o|^2$$
or,
$$ langle z, z_1-z_o rangle = frac{1}{2}left(|z_1|^2- |z_o|^2right) $$
Now, if we have $|z_o| = |z_1|$ then it follows:
$$ langle z, z_1-z_o rangle = 0 $$
which says $z$ points in a direction perpendicular to the line-segment from $z_0$ to $z_1$. In any event, you can use the things I've done here to find the answer.
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
$endgroup$
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
add a comment |
$begingroup$
Begin by noting that $langle x+iy,a+ib rangle = xa+yb = text{Re}left((x+iy)(a-ib)right)$ so if we think of $z$ and $w$ as two-dimensional vectors then the complex algebra permits a natural formulation of the Euclidean inner-product in the plane:
$$ langle v,w rangle = text{Re} (v bar{w}) $$
On the other hand, we have $|z|^2 = z bar{z}$ and the many wonderful properties of the complex conjugate. Suppose $|z-z_o| = rho|z-z_1|$ for some $rho>0$ (if I say $rho>0$ then it must be real !). Square the given condition and use properties of complex conjugation:
$$ (z-z_o)(bar{z}-bar{z_o}) = rho^2(z-z_1)(bar{z}-bar{z_1})$$
or
$$ zbar{z}-zbar{z_o}-z_obar{z}+z_obar{z_o} = rho^2left( zbar{z}-zbar{z_1}-z_1bar{z}+z_1bar{z_1}right)$$
which gives
$$ |z|^2-zbar{z_o}-z_obar{z}+|z_o|^2 = rho^2left( |z|^2-zbar{z_1}-z_1bar{z}+|z_1|^2right)$$
which we can rearrange to
$$ rho^2zbar{z_1}+rho^2z_1bar{z}-zbar{z_o}-z_obar{z} =
(rho^2-1)|z|^2 +rho^2|z_1|^2- |z_o|^2$$
Ok, I'm tired of $rho$, I'm setting $rho=1$ in which case:
$$ zbar{z_1}+z_1bar{z}-zbar{z_o}-z_obar{z} =
|z_1|^2- |z_o|^2$$
which gives, using $text{Re}(z) = frac{1}{2}(z + bar{z})$,
$$ z(overline{z_1-z_o})+bar{z}(z_1-z_o) = 2text{Re}left(z(overline{z_1-z_o})right) = |z_1|^2- |z_o|^2$$
or,
$$ langle z, z_1-z_o rangle = frac{1}{2}left(|z_1|^2- |z_o|^2right) $$
Now, if we have $|z_o| = |z_1|$ then it follows:
$$ langle z, z_1-z_o rangle = 0 $$
which says $z$ points in a direction perpendicular to the line-segment from $z_0$ to $z_1$. In any event, you can use the things I've done here to find the answer.
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
$endgroup$
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
add a comment |
$begingroup$
Begin by noting that $langle x+iy,a+ib rangle = xa+yb = text{Re}left((x+iy)(a-ib)right)$ so if we think of $z$ and $w$ as two-dimensional vectors then the complex algebra permits a natural formulation of the Euclidean inner-product in the plane:
$$ langle v,w rangle = text{Re} (v bar{w}) $$
On the other hand, we have $|z|^2 = z bar{z}$ and the many wonderful properties of the complex conjugate. Suppose $|z-z_o| = rho|z-z_1|$ for some $rho>0$ (if I say $rho>0$ then it must be real !). Square the given condition and use properties of complex conjugation:
$$ (z-z_o)(bar{z}-bar{z_o}) = rho^2(z-z_1)(bar{z}-bar{z_1})$$
or
$$ zbar{z}-zbar{z_o}-z_obar{z}+z_obar{z_o} = rho^2left( zbar{z}-zbar{z_1}-z_1bar{z}+z_1bar{z_1}right)$$
which gives
$$ |z|^2-zbar{z_o}-z_obar{z}+|z_o|^2 = rho^2left( |z|^2-zbar{z_1}-z_1bar{z}+|z_1|^2right)$$
which we can rearrange to
$$ rho^2zbar{z_1}+rho^2z_1bar{z}-zbar{z_o}-z_obar{z} =
(rho^2-1)|z|^2 +rho^2|z_1|^2- |z_o|^2$$
Ok, I'm tired of $rho$, I'm setting $rho=1$ in which case:
$$ zbar{z_1}+z_1bar{z}-zbar{z_o}-z_obar{z} =
|z_1|^2- |z_o|^2$$
which gives, using $text{Re}(z) = frac{1}{2}(z + bar{z})$,
$$ z(overline{z_1-z_o})+bar{z}(z_1-z_o) = 2text{Re}left(z(overline{z_1-z_o})right) = |z_1|^2- |z_o|^2$$
or,
$$ langle z, z_1-z_o rangle = frac{1}{2}left(|z_1|^2- |z_o|^2right) $$
Now, if we have $|z_o| = |z_1|$ then it follows:
$$ langle z, z_1-z_o rangle = 0 $$
which says $z$ points in a direction perpendicular to the line-segment from $z_0$ to $z_1$. In any event, you can use the things I've done here to find the answer.
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
$endgroup$
Begin by noting that $langle x+iy,a+ib rangle = xa+yb = text{Re}left((x+iy)(a-ib)right)$ so if we think of $z$ and $w$ as two-dimensional vectors then the complex algebra permits a natural formulation of the Euclidean inner-product in the plane:
$$ langle v,w rangle = text{Re} (v bar{w}) $$
On the other hand, we have $|z|^2 = z bar{z}$ and the many wonderful properties of the complex conjugate. Suppose $|z-z_o| = rho|z-z_1|$ for some $rho>0$ (if I say $rho>0$ then it must be real !). Square the given condition and use properties of complex conjugation:
$$ (z-z_o)(bar{z}-bar{z_o}) = rho^2(z-z_1)(bar{z}-bar{z_1})$$
or
$$ zbar{z}-zbar{z_o}-z_obar{z}+z_obar{z_o} = rho^2left( zbar{z}-zbar{z_1}-z_1bar{z}+z_1bar{z_1}right)$$
which gives
$$ |z|^2-zbar{z_o}-z_obar{z}+|z_o|^2 = rho^2left( |z|^2-zbar{z_1}-z_1bar{z}+|z_1|^2right)$$
which we can rearrange to
$$ rho^2zbar{z_1}+rho^2z_1bar{z}-zbar{z_o}-z_obar{z} =
(rho^2-1)|z|^2 +rho^2|z_1|^2- |z_o|^2$$
Ok, I'm tired of $rho$, I'm setting $rho=1$ in which case:
$$ zbar{z_1}+z_1bar{z}-zbar{z_o}-z_obar{z} =
|z_1|^2- |z_o|^2$$
which gives, using $text{Re}(z) = frac{1}{2}(z + bar{z})$,
$$ z(overline{z_1-z_o})+bar{z}(z_1-z_o) = 2text{Re}left(z(overline{z_1-z_o})right) = |z_1|^2- |z_o|^2$$
or,
$$ langle z, z_1-z_o rangle = frac{1}{2}left(|z_1|^2- |z_o|^2right) $$
Now, if we have $|z_o| = |z_1|$ then it follows:
$$ langle z, z_1-z_o rangle = 0 $$
which says $z$ points in a direction perpendicular to the line-segment from $z_0$ to $z_1$. In any event, you can use the things I've done here to find the answer.
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
answered Aug 5 '15 at 20:15
James S. CookJames S. Cook
13.1k22872
13.1k22872
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
add a comment |
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
$begingroup$
ok, so while I have not answered the question I at least showed you why $rho=1$ is not allowed.
$endgroup$
– James S. Cook
Aug 5 '15 at 20:20
add a comment |
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$begingroup$
Why don't you tackle it geometrically? Draw a line from $z_0$ to $z_1$, draw the points on this line satisfying the equation (there must be two), then find the centre point between these two, the equation of that circle, and show all points satisfy the equation?
$endgroup$
– Colm Bhandal
Aug 5 '15 at 19:41
$begingroup$
Then I have to show that no other point satisfy that eqn. Will you do it elaborately and give me the answer? Because I think doing algebraic way will be shorter.
$endgroup$
– user152715
Aug 5 '15 at 19:44
$begingroup$
Have you already treated Möbius transformations?
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– Daniel Fischer♦
Aug 5 '15 at 19:50
$begingroup$
No, I have not used it. This is first chapter problem in Gamelin's book and I want to do it in an elementary way.
$endgroup$
– user152715
Aug 5 '15 at 19:52
1
$begingroup$
Okay, then do it per pedes. So you found that the condition is equivalent to $alvert zrvert^2 - 2 operatorname{Re}(boverline{z}) + c = 0$ for some real $a,c$ with $a neq 0$ and $b in mathbb{C}$. When is that sort of equation the defining equation for a circle?
$endgroup$
– Daniel Fischer♦
Aug 5 '15 at 19:55