The minimum value of $|z-1+2i| + |4i-3-z|$ is [closed]
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
closed as off-topic by max_zorn, rtybase, TheSimpliFire, Holo, user21820 Nov 25 '18 at 9:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – max_zorn, rtybase, TheSimpliFire, Holo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
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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
closed as off-topic by max_zorn, rtybase, TheSimpliFire, Holo, user21820 Nov 25 '18 at 9:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – max_zorn, rtybase, TheSimpliFire, Holo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
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
Nov 19 '18 at 5:21
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
Nov 19 '18 at 5:52
add a comment |
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
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 Nov 19 '18 at 8:17
jayant98
472115
472115
asked Nov 19 '18 at 5:14
Samarth Mankan
11
11
closed as off-topic by max_zorn, rtybase, TheSimpliFire, Holo, user21820 Nov 25 '18 at 9:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – max_zorn, rtybase, TheSimpliFire, Holo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by max_zorn, rtybase, TheSimpliFire, Holo, user21820 Nov 25 '18 at 9:44
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – max_zorn, rtybase, TheSimpliFire, Holo, user21820
If this question can be reworded to fit the rules in the help center, please edit the question.
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
Nov 19 '18 at 5:21
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
Nov 19 '18 at 5:52
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
Nov 19 '18 at 5:21
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
Nov 19 '18 at 5:52
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
Nov 19 '18 at 5:21
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
Nov 19 '18 at 5:21
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
Nov 19 '18 at 5:52
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
Nov 19 '18 at 5:52
add a comment |
4 Answers
4
active
oldest
votes
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
add a comment |
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
add a comment |
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
add a comment |
A bit of geometry in the complex plane:
1)$d:=$
$|z-(1-2i)| +|z-(-3+4i)|$ , $z=x+iy$.
$A(1,-2i)$, $B(-3,4i)$, $C(x,iy)$.
1) $A,B,C$ are not collinear.
In $triangle ABC:$
$d= |AC|+|BC| >|AB|.
(Strict triangle inequality ).
2) $A,B,C$ are collinear.
a) $z$ is within the line segment $AB$,
then $d=|AB|$((why?).
b) $z$ is outside the line segment $|AB|$,
then $d>|AB|$(why?).
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
add a comment |
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
add a comment |
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
Hint: The sum of two distances of a point $z$ from two points is minimum when $z$ is between them.
answered Nov 19 '18 at 5:19
Nosrati
26.5k62353
26.5k62353
add a comment |
add a comment |
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
add a comment |
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
add a comment |
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
The minimum is the distance between $1-2i$ and $3-4i$ which is 4$sqrt {(3-1)^2+(-4+2)^2} = 2sqrt 2 $$
This is when $z$ is on the segment joining the two points and $z$ is between them.
answered Nov 19 '18 at 5:51
Mohammad Riazi-Kermani
40.8k42059
40.8k42059
add a comment |
add a comment |
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
add a comment |
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
add a comment |
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|z-a| + |z-b| geq |z-a - (z-b)| = |b-a| = |-4 +6i| = 2sqrt{13}$$
Note, that the minimum is attained when $z$ lies on the segment connecting $a$ and $b$.
You may proceed as follows:
You have
$|z-a| + |z-b| stackrel{!}{rightarrow} mbox{Min}$ with
$a= 1-2i$ and $b = -3+4i$
The triangle inequality gives immediately
$$|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 $b$.
edited Nov 19 '18 at 8:00
answered Nov 19 '18 at 7:53
trancelocation
9,1051521
9,1051521
add a comment |
add a comment |
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?).
add a comment |
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?).
add a comment |
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?).
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?).
edited Nov 19 '18 at 10:06
answered Nov 19 '18 at 9:48
Peter Szilas
10.7k2720
10.7k2720
add a comment |
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
Nov 19 '18 at 5:21
An equation whose solution set is the segment $overline{AB}$ is $|A-P|+|P-B| = |A-B|$
– steven gregory
Nov 19 '18 at 5:52