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The minimum value of $|z-1+2i| + |4i-3-z|$ is
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The minimum value of $$|z-1+2i| + |4i-3-z|$$ is?
The only method of moving further that comes to my mind is assuming $$z=x+iy$$.
algebra-precalculus complex-numbers
edited yesterday
jayant98
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Samarth Mankan
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down vote
favorite
The minimum value of $$|z-1+2i| + |4i-3-z|$$ is?
The only method of moving further that comes to my mind is assuming $$z=x+iy$$.
algebra-precalculus complex-numbers
edited yesterday
jayant98
12613
asked yesterday
Samarth Mankan
11
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is a reasonable approach and the one I use if I don't have a better idea. What happened when you tried it? You have a function of two real variables, compute it, take the derivatives, set to zero, and what happens? There is an easier geometric approach if you think about what the absolute values represent.
– Ross Millikan
yesterday
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The minimum value of $$|z-1+2i| + |4i-3-z|$$ is?
The only method of moving further that comes to my mind is assuming $$z=x+iy$$.
algebra-precalculus complex-numbers
edited yesterday
jayant98
12613
asked yesterday
Samarth Mankan
11
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The minimum value of $$|z-1+2i| + |4i-3-z|$$ is?
The only method of moving further that comes to my mind is assuming $$z=x+iy$$.
algebra-precalculus complex-numbers
algebra-precalculus complex-numbers
edited yesterday
jayant98
12613
asked yesterday
Samarth Mankan
11
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
jayant98
12613
asked yesterday
Samarth Mankan
11
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
jayant98
12613
edited yesterday
jayant98
12613
edited yesterday
jayant98
12613
12613
asked yesterday
Samarth Mankan
11
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Samarth Mankan
11
asked yesterday
Samarth Mankan
11
11
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Samarth Mankan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
That is a reasonable approach and the one I use if I don't have a better idea. What happened when you tried it? You have a function of two real variables, compute it, take the derivatives, set to zero, and what happens? There is an easier geometric approach if you think about what the absolute values represent.
– Ross Millikan
yesterday
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
yesterday
add a comment |
That is a reasonable approach and the one I use if I don't have a better idea. What happened when you tried it? You have a function of two real variables, compute it, take the derivatives, set to zero, and what happens? There is an easier geometric approach if you think about what the absolute values represent.
– Ross Millikan
yesterday
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
yesterday
That is a reasonable approach and the one I use if I don't have a better idea. What happened when you tried it? You have a function of two real variables, compute it, take the derivatives, set to zero, and what happens? There is an easier geometric approach if you think about what the absolute values represent.
– Ross Millikan
yesterday
That is a reasonable approach and the one I use if I don't have a better idea. What happened when you tried it? You have a function of two real variables, compute it, take the derivatives, set to zero, and what happens? There is an easier geometric approach if you think about what the absolute values represent.
– Ross Millikan
yesterday
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
yesterday
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
yesterday
add a comment |
4 Answers
4
active
oldest
votes
up vote
3
down vote
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
answered yesterday
Nosrati
25.8k62252
add a comment |
up vote
0
down vote
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
add a comment |
up vote
0
down vote
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
0
down vote
A bit of geometry in the complex plane:
1)$d:=$
$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.
$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.
1) $A,B,C$ are not collinear.
In $triangle ABC:$
$d= |AC|+|BC| >|AB|.
(Strict triangle inequality ).
2) $A,B,C$ are collinear.
a) $z$ is within the line segment $AB$,
then $d=|AB|$((why?).
b) $z$ is outside the line segment $|AB|$,
then $d>|AB|$(why?).
answered yesterday
Peter Szilas
9,9292720
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
answered yesterday
Nosrati
25.8k62252
add a comment |
up vote
3
down vote
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
answered yesterday
Nosrati
25.8k62252
add a comment |
up vote
3
down vote
up vote
3
down vote
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
answered yesterday
Nosrati
25.8k62252
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
answered yesterday
Nosrati
25.8k62252
answered yesterday
Nosrati
25.8k62252
answered yesterday
Nosrati
25.8k62252
answered yesterday
Nosrati
25.8k62252
25.8k62252
add a comment |
add a comment |
up vote
0
down vote
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
add a comment |
up vote
0
down vote
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
add a comment |
up vote
0
down vote
up vote
0
down vote
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
answered yesterday
Mohammad Riazi-Kermani
40.2k41958
40.2k41958
add a comment |
add a comment |
up vote
0
down vote
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
0
down vote
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
answered yesterday
trancelocation
8,0561519
add a comment |
up vote
0
down vote
up vote
0
down vote
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
answered yesterday
trancelocation
8,0561519
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
answered yesterday
trancelocation
8,0561519
edited yesterday
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
answered yesterday
trancelocation
8,0561519
8,0561519
add a comment |
add a comment |
up vote
0
down vote
A bit of geometry in the complex plane:
1)$d:=$
$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.
$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.
1) $A,B,C$ are not collinear.
In $triangle ABC:$
$d= |AC|+|BC| >|AB|.
(Strict triangle inequality ).
2) $A,B,C$ are collinear.
a) $z$ is within the line segment $AB$,
then $d=|AB|$((why?).
b) $z$ is outside the line segment $|AB|$,
then $d>|AB|$(why?).
answered yesterday
Peter Szilas
9,9292720
add a comment |
up vote
0
down vote
A bit of geometry in the complex plane:
1)$d:=$
$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.
$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.
1) $A,B,C$ are not collinear.
In $triangle ABC:$
$d= |AC|+|BC| >|AB|.
(Strict triangle inequality ).
2) $A,B,C$ are collinear.
a) $z$ is within the line segment $AB$,
then $d=|AB|$((why?).
b) $z$ is outside the line segment $|AB|$,
then $d>|AB|$(why?).
answered yesterday
Peter Szilas
9,9292720
add a comment |
up vote
0
down vote
up vote
0
down vote
A bit of geometry in the complex plane:
1)$d:=$
$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.
$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.
1) $A,B,C$ are not collinear.
In $triangle ABC:$
$d= |AC|+|BC| >|AB|.
(Strict triangle inequality ).
2) $A,B,C$ are collinear.
a) $z$ is within the line segment $AB$,
then $d=|AB|$((why?).
b) $z$ is outside the line segment $|AB|$,
then $d>|AB|$(why?).
answered yesterday
Peter Szilas
9,9292720
A bit of geometry in the complex plane:
1)$d:=$
$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.
$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.
1) $A,B,C$ are not collinear.
In $triangle ABC:$
$d= |AC|+|BC| >|AB|.
(Strict triangle inequality ).
2) $A,B,C$ are collinear.
a) $z$ is within the line segment $AB$,
then $d=|AB|$((why?).
b) $z$ is outside the line segment $|AB|$,
then $d>|AB|$(why?).
answered yesterday
Peter Szilas
9,9292720
edited yesterday
answered yesterday
Peter Szilas
9,9292720
answered yesterday
Peter Szilas
9,9292720
answered yesterday
Peter Szilas
9,9292720
9,9292720
add a comment |
add a comment |
Samarth Mankan is a new contributor. Be nice, and check out our Code of Conduct.
Samarth Mankan is a new contributor. Be nice, and check out our Code of Conduct.
Samarth Mankan is a new contributor. Be nice, and check out our Code of Conduct.
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That is a reasonable approach and the one I use if I don't have a better idea. What happened when you tried it? You have a function of two real variables, compute it, take the derivatives, set to zero, and what happens? There is an easier geometric approach if you think about what the absolute values represent.
– Ross Millikan
yesterday
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
yesterday