Find symmetric points with respect to the unit circle
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I have struggled with an exercise, namely: find the set of symmetric points with respect to the unit circle of a circle given by this equation: $ |z-1|=1$, I have an idea of what this might be. Since points, A and B, which are the intersection points of both circles will not change when taking symmetry and also point 2 will go to 1/2. So I guess that the set will be the line crossing all 3 points A, B and 1/2, but I would like some explanation added to this. Thanks, any help appreciated.
complex-numbers complex-geometry
asked Nov 21 at 16:11
ryszard eggink
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up vote
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down vote
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I have struggled with an exercise, namely: find the set of symmetric points with respect to the unit circle of a circle given by this equation: $ |z-1|=1$, I have an idea of what this might be. Since points, A and B, which are the intersection points of both circles will not change when taking symmetry and also point 2 will go to 1/2. So I guess that the set will be the line crossing all 3 points A, B and 1/2, but I would like some explanation added to this. Thanks, any help appreciated.
complex-numbers complex-geometry
asked Nov 21 at 16:11
ryszard eggink
1369
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have struggled with an exercise, namely: find the set of symmetric points with respect to the unit circle of a circle given by this equation: $ |z-1|=1$, I have an idea of what this might be. Since points, A and B, which are the intersection points of both circles will not change when taking symmetry and also point 2 will go to 1/2. So I guess that the set will be the line crossing all 3 points A, B and 1/2, but I would like some explanation added to this. Thanks, any help appreciated.
complex-numbers complex-geometry
asked Nov 21 at 16:11
ryszard eggink
1369
I have struggled with an exercise, namely: find the set of symmetric points with respect to the unit circle of a circle given by this equation: $ |z-1|=1$, I have an idea of what this might be. Since points, A and B, which are the intersection points of both circles will not change when taking symmetry and also point 2 will go to 1/2. So I guess that the set will be the line crossing all 3 points A, B and 1/2, but I would like some explanation added to this. Thanks, any help appreciated.
complex-numbers complex-geometry
complex-numbers complex-geometry
asked Nov 21 at 16:11
ryszard eggink
1369
asked Nov 21 at 16:11
ryszard eggink
1369
asked Nov 21 at 16:11
ryszard eggink
1369
asked Nov 21 at 16:11
ryszard eggink
1369
asked Nov 21 at 16:11
ryszard eggink
1369
1369
add a comment |
add a comment |
1 Answer
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Hint: Let $z$ and $w$ be on circle $|z-1|=1$, then
$$|z-1|=1~~~~;~~~~|w-1|=1$$
these points are symmetric respct to the unit circle, means there is a $thetainmathbb R$ such that
$$dfrac{z+w}{2}=e^{itheta}$$
or $|z+w|=2$, therefore with deleting $w$ among equations
$$|z-1|=1~~~~;~~~~|w-1|=1~~~~;~~~~|z+w|=2$$
we can find desired points $z$.
answered Nov 21 at 19:07
Nosrati
1
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: Let $z$ and $w$ be on circle $|z-1|=1$, then
$$|z-1|=1~~~~;~~~~|w-1|=1$$
these points are symmetric respct to the unit circle, means there is a $thetainmathbb R$ such that
$$dfrac{z+w}{2}=e^{itheta}$$
or $|z+w|=2$, therefore with deleting $w$ among equations
$$|z-1|=1~~~~;~~~~|w-1|=1~~~~;~~~~|z+w|=2$$
we can find desired points $z$.
answered Nov 21 at 19:07
Nosrati
1
add a comment |
up vote
0
down vote
Hint: Let $z$ and $w$ be on circle $|z-1|=1$, then
$$|z-1|=1~~~~;~~~~|w-1|=1$$
these points are symmetric respct to the unit circle, means there is a $thetainmathbb R$ such that
$$dfrac{z+w}{2}=e^{itheta}$$
or $|z+w|=2$, therefore with deleting $w$ among equations
$$|z-1|=1~~~~;~~~~|w-1|=1~~~~;~~~~|z+w|=2$$
we can find desired points $z$.
answered Nov 21 at 19:07
Nosrati
1
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint: Let $z$ and $w$ be on circle $|z-1|=1$, then
$$|z-1|=1~~~~;~~~~|w-1|=1$$
these points are symmetric respct to the unit circle, means there is a $thetainmathbb R$ such that
$$dfrac{z+w}{2}=e^{itheta}$$
or $|z+w|=2$, therefore with deleting $w$ among equations
$$|z-1|=1~~~~;~~~~|w-1|=1~~~~;~~~~|z+w|=2$$
we can find desired points $z$.
answered Nov 21 at 19:07
Nosrati
1
Hint: Let $z$ and $w$ be on circle $|z-1|=1$, then
$$|z-1|=1~~~~;~~~~|w-1|=1$$
these points are symmetric respct to the unit circle, means there is a $thetainmathbb R$ such that
$$dfrac{z+w}{2}=e^{itheta}$$
or $|z+w|=2$, therefore with deleting $w$ among equations
$$|z-1|=1~~~~;~~~~|w-1|=1~~~~;~~~~|z+w|=2$$
we can find desired points $z$.
answered Nov 21 at 19:07
Nosrati
1
edited Nov 21 at 19:14
answered Nov 21 at 19:07
Nosrati
1
answered Nov 21 at 19:07
Nosrati
1
answered Nov 21 at 19:07
Nosrati
1
1
add a comment |
add a comment |
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