How to show that $frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 Leftrightarrow |z|leq 1$ [closed]
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I can't solve this math problem and I'm stuck in it! I think it must have a trick and I don't know that. the problem is this:
if u and v and z are complex numbers and $|u|<1$ and $|v|=1$ show that:
$$
frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 Leftrightarrow |z|leq 1
$$
any help or tip to solve it would be appreciated.
inequality complex-numbers
edited Nov 24 at 14:54
Nosrati
26.1k62353
asked Nov 24 at 14:48
Peyman
165
closed as off-topic by Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus Nov 26 at 2:54
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." – Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-1
down vote
favorite
I can't solve this math problem and I'm stuck in it! I think it must have a trick and I don't know that. the problem is this:
if u and v and z are complex numbers and $|u|<1$ and $|v|=1$ show that:
$$
frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 Leftrightarrow |z|leq 1
$$
any help or tip to solve it would be appreciated.
inequality complex-numbers
edited Nov 24 at 14:54
Nosrati
26.1k62353
asked Nov 24 at 14:48
Peyman
165
closed as off-topic by Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus Nov 26 at 2:54
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." – Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
It might be beneficial to move the denominator to the other side, and then square both sides.
– Cameron Williams
Nov 24 at 14:51
@CameronWilliams I've done that but it made it more complicated.
– Peyman
Nov 24 at 14:55
See math.stackexchange.com/questions/1410972
– Nosrati
Nov 24 at 14:59
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I can't solve this math problem and I'm stuck in it! I think it must have a trick and I don't know that. the problem is this:
if u and v and z are complex numbers and $|u|<1$ and $|v|=1$ show that:
$$
frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 Leftrightarrow |z|leq 1
$$
any help or tip to solve it would be appreciated.
inequality complex-numbers
edited Nov 24 at 14:54
Nosrati
26.1k62353
asked Nov 24 at 14:48
Peyman
165
I can't solve this math problem and I'm stuck in it! I think it must have a trick and I don't know that. the problem is this:
if u and v and z are complex numbers and $|u|<1$ and $|v|=1$ show that:
$$
frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 Leftrightarrow |z|leq 1
$$
any help or tip to solve it would be appreciated.
inequality complex-numbers
inequality complex-numbers
edited Nov 24 at 14:54
Nosrati
26.1k62353
asked Nov 24 at 14:48
Peyman
165
edited Nov 24 at 14:54
Nosrati
26.1k62353
asked Nov 24 at 14:48
Peyman
165
edited Nov 24 at 14:54
Nosrati
26.1k62353
edited Nov 24 at 14:54
Nosrati
26.1k62353
edited Nov 24 at 14:54
Nosrati
26.1k62353
26.1k62353
asked Nov 24 at 14:48
Peyman
165
asked Nov 24 at 14:48
Peyman
165
asked Nov 24 at 14:48
Peyman
165
165
closed as off-topic by Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus Nov 26 at 2:54
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." – Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus Nov 26 at 2:54
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." – Nosrati, Jyrki Lahtonen, user302797, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
It might be beneficial to move the denominator to the other side, and then square both sides.
– Cameron Williams
Nov 24 at 14:51
@CameronWilliams I've done that but it made it more complicated.
– Peyman
Nov 24 at 14:55
See math.stackexchange.com/questions/1410972
– Nosrati
Nov 24 at 14:59
add a comment |
It might be beneficial to move the denominator to the other side, and then square both sides.
– Cameron Williams
Nov 24 at 14:51
@CameronWilliams I've done that but it made it more complicated.
– Peyman
Nov 24 at 14:55
See math.stackexchange.com/questions/1410972
– Nosrati
Nov 24 at 14:59
It might be beneficial to move the denominator to the other side, and then square both sides.
– Cameron Williams
Nov 24 at 14:51
It might be beneficial to move the denominator to the other side, and then square both sides.
– Cameron Williams
Nov 24 at 14:51
@CameronWilliams I've done that but it made it more complicated.
– Peyman
Nov 24 at 14:55
@CameronWilliams I've done that but it made it more complicated.
– Peyman
Nov 24 at 14:55
See math.stackexchange.com/questions/1410972
– Nosrati
Nov 24 at 14:59
See math.stackexchange.com/questions/1410972
– Nosrati
Nov 24 at 14:59
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
HINT
We have that
$$frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 iff frac{|u - bar{v} z|^2}{|bar{u} z - v|^2} leq 1 iff |u - bar{v} z|^2le |bar{u} z - v|^2$$
$$(u - bar{v} z)(bar u - v bar z)le(bar{u} z - v)(ubar{z} - bar v)$$
$$|u|^2-uvbar z-bar u bar v z+|v|^2|z|^2le|u|^2|z|^2-bar u bar v z-uvbar z+|v|^2$$
answered Nov 24 at 15:00
gimusi
89.4k74495
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
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
accepted
HINT
We have that
$$frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 iff frac{|u - bar{v} z|^2}{|bar{u} z - v|^2} leq 1 iff |u - bar{v} z|^2le |bar{u} z - v|^2$$
$$(u - bar{v} z)(bar u - v bar z)le(bar{u} z - v)(ubar{z} - bar v)$$
$$|u|^2-uvbar z-bar u bar v z+|v|^2|z|^2le|u|^2|z|^2-bar u bar v z-uvbar z+|v|^2$$
answered Nov 24 at 15:00
gimusi
89.4k74495
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
add a comment |
up vote
0
down vote
accepted
HINT
We have that
$$frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 iff frac{|u - bar{v} z|^2}{|bar{u} z - v|^2} leq 1 iff |u - bar{v} z|^2le |bar{u} z - v|^2$$
$$(u - bar{v} z)(bar u - v bar z)le(bar{u} z - v)(ubar{z} - bar v)$$
$$|u|^2-uvbar z-bar u bar v z+|v|^2|z|^2le|u|^2|z|^2-bar u bar v z-uvbar z+|v|^2$$
answered Nov 24 at 15:00
gimusi
89.4k74495
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
HINT
We have that
$$frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 iff frac{|u - bar{v} z|^2}{|bar{u} z - v|^2} leq 1 iff |u - bar{v} z|^2le |bar{u} z - v|^2$$
$$(u - bar{v} z)(bar u - v bar z)le(bar{u} z - v)(ubar{z} - bar v)$$
$$|u|^2-uvbar z-bar u bar v z+|v|^2|z|^2le|u|^2|z|^2-bar u bar v z-uvbar z+|v|^2$$
answered Nov 24 at 15:00
gimusi
89.4k74495
HINT
We have that
$$frac{|u - bar{v} z|}{|bar{u} z - v|} leq 1 iff frac{|u - bar{v} z|^2}{|bar{u} z - v|^2} leq 1 iff |u - bar{v} z|^2le |bar{u} z - v|^2$$
$$(u - bar{v} z)(bar u - v bar z)le(bar{u} z - v)(ubar{z} - bar v)$$
$$|u|^2-uvbar z-bar u bar v z+|v|^2|z|^2le|u|^2|z|^2-bar u bar v z-uvbar z+|v|^2$$
answered Nov 24 at 15:00
gimusi
89.4k74495
answered Nov 24 at 15:00
gimusi
89.4k74495
answered Nov 24 at 15:00
gimusi
89.4k74495
answered Nov 24 at 15:00
gimusi
89.4k74495
89.4k74495
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
add a comment |
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
OMG! I get it. thanks
– Peyman
Nov 24 at 15:13
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
@Peyman That's nice, recall that conjugating is a key property to deal with complex equations or inequalities.
– gimusi
Nov 24 at 15:14
add a comment |
It might be beneficial to move the denominator to the other side, and then square both sides.
– Cameron Williams
Nov 24 at 14:51
@CameronWilliams I've done that but it made it more complicated.
– Peyman
Nov 24 at 14:55
See math.stackexchange.com/questions/1410972
– Nosrati
Nov 24 at 14:59