Finding minimum value of the equation given.
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Question - Find the minimum value of $|1 + z| + |1-z|$.
I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.
complex-analysis complex-numbers
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up vote
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Question - Find the minimum value of $|1 + z| + |1-z|$.
I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.
complex-analysis complex-numbers
$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42
@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Question - Find the minimum value of $|1 + z| + |1-z|$.
I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.
complex-analysis complex-numbers
Question - Find the minimum value of $|1 + z| + |1-z|$.
I'm trying to solve the question by thinking of them as points in the Argand plane.
The $|1+z|$ can be written as $|z - (-1)|$ which is the distance of $z$ from $(-1) $ on the Argand plane.
But I don't understand how to find the second part on Argand plane like I did the first one. If I find the second point, then the answer will just be the minimum distance between both the points.
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Nov 17 at 17:45
Ted Shifrin
62.1k44489
62.1k44489
asked Nov 16 at 12:32
Kaustuv Sawarn
515
515
$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42
@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47
add a comment |
$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42
@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47
$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42
$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42
@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47
@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.
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$|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.
BEcause the axis must be longer than the distance between focus, $cge 2$.
add a comment |
up vote
1
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A bit of geometry.
In the complex plane:
$A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.
$triangle ABC$ has sides of lengths:
$|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.
The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:
$|z+1|+|z-1| > 2.$
$2$ is a lower bound. Is there a minimum?
If yes , $z=?$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.
add a comment |
up vote
1
down vote
accepted
The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.
The formula represents the sum of the distances from z to 1 and -1. So the minimum is at the midpoint of -1 , 1, i. e. 0; thus the minimum value is 2.
answered Nov 16 at 14:03
i. m. soloveichik
3,68011125
3,68011125
add a comment |
add a comment |
up vote
1
down vote
$|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.
BEcause the axis must be longer than the distance between focus, $cge 2$.
add a comment |
up vote
1
down vote
$|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.
BEcause the axis must be longer than the distance between focus, $cge 2$.
add a comment |
up vote
1
down vote
up vote
1
down vote
$|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.
BEcause the axis must be longer than the distance between focus, $cge 2$.
$|1+z|+|1-z|=c$ for a positive constant $c>0$ dscribes an ellipse with focus at $1$ and $-1$ and axis of length $c$.
BEcause the axis must be longer than the distance between focus, $cge 2$.
answered Nov 16 at 13:24
Tito Eliatron
1,012621
1,012621
add a comment |
add a comment |
up vote
1
down vote
A bit of geometry.
In the complex plane:
$A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.
$triangle ABC$ has sides of lengths:
$|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.
The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:
$|z+1|+|z-1| > 2.$
$2$ is a lower bound. Is there a minimum?
If yes , $z=?$
add a comment |
up vote
1
down vote
A bit of geometry.
In the complex plane:
$A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.
$triangle ABC$ has sides of lengths:
$|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.
The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:
$|z+1|+|z-1| > 2.$
$2$ is a lower bound. Is there a minimum?
If yes , $z=?$
add a comment |
up vote
1
down vote
up vote
1
down vote
A bit of geometry.
In the complex plane:
$A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.
$triangle ABC$ has sides of lengths:
$|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.
The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:
$|z+1|+|z-1| > 2.$
$2$ is a lower bound. Is there a minimum?
If yes , $z=?$
A bit of geometry.
In the complex plane:
$A(-1,0)$, $B(1,0)$ and let $C(x,y),$ where $z =x+iy$, $x,y$, real.
$triangle ABC$ has sides of lengths:
$|AB|=2$, |$AC| =|z+1|$, $|BC|= |z-1|$.
The sum of the lengths of 2 sides of a triangle is greater than the 3rd side:
$|z+1|+|z-1| > 2.$
$2$ is a lower bound. Is there a minimum?
If yes , $z=?$
edited Nov 16 at 13:38
answered Nov 16 at 13:29
Peter Szilas
10.1k2720
10.1k2720
add a comment |
add a comment |
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$|1-z| = |-(z-1)|=|z-1|$
– krirkrirk
Nov 16 at 12:42
@krirkrirk Thanks! It didn't click at the moment. Got the answer!
– Kaustuv Sawarn
Nov 16 at 12:47