Greatest value of $|z|$ given that $left|z- frac 4z right| = 2$. Why does using triangle inequality in this...
$begingroup$
Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?
This has been asked here before but my question is about a specific method that doesn't seem to work.
The method is:
$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$
$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)
which is true for all values of $|z|$.
Hence $|z| in (0, infty)$.
Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?
A user has also asked that in one of the comments but hasn't received a reply there.
algebra-precalculus proof-verification
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
$endgroup$
add a comment |
$begingroup$
Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?
This has been asked here before but my question is about a specific method that doesn't seem to work.
The method is:
$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$
$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)
which is true for all values of $|z|$.
Hence $|z| in (0, infty)$.
Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?
A user has also asked that in one of the comments but hasn't received a reply there.
algebra-precalculus proof-verification
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
$endgroup$
$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50
$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51
1
$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52
1
$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59
add a comment |
$begingroup$
Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?
This has been asked here before but my question is about a specific method that doesn't seem to work.
The method is:
$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$
$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)
which is true for all values of $|z|$.
Hence $|z| in (0, infty)$.
Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?
A user has also asked that in one of the comments but hasn't received a reply there.
algebra-precalculus proof-verification
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
$endgroup$
Greatest value of $|z|$ given that $left|z- dfrac 4z right| = 2$ is?
This has been asked here before but my question is about a specific method that doesn't seem to work.
The method is:
$left|z- dfrac 4z right|le |z|+ left|dfrac{4}{z}right|implies 2 le |z|+ dfrac{4}{|z|}$
$implies |z|^2 - 2|z| +4ge 0$ since $|z| > 0$ (here)
which is true for all values of $|z|$.
Hence $|z| in (0, infty)$.
Then why is the maximum given by $|z| =1 + sqrt 5$, in other words, what's the fault in this method ?
A user has also asked that in one of the comments but hasn't received a reply there.
algebra-precalculus proof-verification
algebra-precalculus proof-verification
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
edited Dec 12 '18 at 21:30
Lorenzo B.
1,8402520
1,8402520
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
asked Dec 12 '18 at 20:45
AbcdAbcd
3,03831235
3,03831235
$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50
$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51
1
$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52
1
$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59
add a comment |
$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50
$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51
1
$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52
1
$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59
$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50
$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50
$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51
$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51
1
1
$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52
$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52
1
1
$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59
$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can apply the triangle inequality this way:
$2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
$endgroup$
add a comment |
$begingroup$
I was solving a similar problem in this question
edit
Assume $zneq 0.;$
$left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.
Solution
- If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
$$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$
Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$
If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.
In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$
The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You can apply the triangle inequality this way:
$2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
$endgroup$
add a comment |
$begingroup$
You can apply the triangle inequality this way:
$2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
$endgroup$
add a comment |
$begingroup$
You can apply the triangle inequality this way:
$2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
$endgroup$
You can apply the triangle inequality this way:
$2 geq |z-frac{4}{z}| geq |z|-|frac{4}{z}|$, whence $|z|^2-2|z|-4 leq 0$ whence $|z| leq 1 + sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+sqrt{5}$
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
answered Dec 12 '18 at 21:05
Sorin TircSorin Tirc
1,755213
1,755213
add a comment |
add a comment |
$begingroup$
I was solving a similar problem in this question
edit
Assume $zneq 0.;$
$left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.
Solution
- If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
$$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$
Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$
If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.
In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$
The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
$endgroup$
add a comment |
$begingroup$
I was solving a similar problem in this question
edit
Assume $zneq 0.;$
$left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.
Solution
- If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
$$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$
Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$
If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.
In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$
The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
$endgroup$
add a comment |
$begingroup$
I was solving a similar problem in this question
edit
Assume $zneq 0.;$
$left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.
Solution
- If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
$$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$
Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$
If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.
In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$
The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
$endgroup$
I was solving a similar problem in this question
edit
Assume $zneq 0.;$
$left|z-frac{4}{z}right|$ is the distance of points representing $z$ and $frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.
Solution
- If $z$ is real, then $z, frac 4z$ lie on the same half-line starting in $0$ and we have
$$left|z-frac 4z right|=|z|-frac {4}{|z|}quad text{or} quad left| z-frac 4z right|=frac{4}{|z|}-|z|,$$
and so $$2=|z|-frac {4}{|z|} quad text{or} quad 2=frac{4}{|z|}-|z|.$$
Multiplying by $|z|$ and solving the quadratic equations gives positive solutions $|z|=sqrt 5 +1;$ from the first and $|z|=sqrt 5 -1;$ from the second one. Note that $sqrt 5 -1=frac{4}{sqrt 5 +1}.$
If $z=ib, bin mathbb{R},$ then $0$ lies on the segment with bounds $z, frac 4z$. The equation to solve is then $2=|z|+frac {4}{|z|}$ and doesn't have solution.
In all other cases, by triangle inequality is $|z|<sqrt 5 +1.$
The maximal value of $|z|$ is $sqrt5 + 1,$ the only convenient numbers are $z=pm(sqrt 5 +1).$
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
edited Dec 13 '18 at 12:11
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
answered Dec 12 '18 at 21:08
user376343user376343
3,6683827
3,6683827
add a comment |
add a comment |
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$begingroup$
the fault of the method is that the triangle inequality gives you something that could be bigger. It doesn't help us to say a hundred million godzillion might be bigger than the maximum.
$endgroup$
– fleablood
Dec 12 '18 at 20:50
$begingroup$
"which is true for all values of $|z|$." Which doesn't tell you anything, because what you've basically done is just confirming that the triangle inequality holds.
$endgroup$
– Arthur
Dec 12 '18 at 20:51
1
$begingroup$
Basically that shows $|z| le 1+sqrt{5} le infty$. that's perfectly true. And you did nothing wrong. It just doesn't help us in the least bit.
$endgroup$
– fleablood
Dec 12 '18 at 20:52
1
$begingroup$
Suppose several people were to guess my brother's weight. One person puts him on a scale and says "He is 178 lbs and 3 oz". Another person weighs him against a cat, then a dog then a sheep then a seal. And says "He weighs between 120 lbs and 250 lbs". A third person weighs him against a battleship and says "He weighs less then 300,000 tons". What was wrong with the battleship method? Technically not a thing.
$endgroup$
– fleablood
Dec 12 '18 at 20:59