Mixing
Showing that $left|fraczz^3+1right|leqfracRR^3-1$
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I am trying to show that if $|z|=R>1$, then
$$left|fraczz^3+1right|leqfracRR^3-1$$
I've been playing around with the above inequality and this is where I am at.
beginalign
textConsider left|z^3+1right|&leqleft|z^3right|+1 text(by the triangle inequality)\
&=left|zright|^3+1 \
&=R^3+1 \ \
Rightarrow left|z^3+1right|&leq R^3+1 \
frac1&geqfrac1R^3+1 \
frac&geqfracR^3+1 textz \
left|fraczz^3+1right|&geqfracRR^3+1 \
endalign
I'm unsure of how to yield the desired inequality. Have I made a mistake somewhere?
EDIT
If I used the inequality $$|z_1+z_2|geq |z_1|-|z_2|$$ I believe this will work. My question is, does this inequality hold for real numbers as well?
complex-analysis proof-verification inequality
asked Aug 3 at 0:53
Bell
560112
add a comment |Â
up vote
2
down vote
favorite
I am trying to show that if $|z|=R>1$, then
$$left|fraczz^3+1right|leqfracRR^3-1$$
I've been playing around with the above inequality and this is where I am at.
beginalign
textConsider left|z^3+1right|&leqleft|z^3right|+1 text(by the triangle inequality)\
&=left|zright|^3+1 \
&=R^3+1 \ \
Rightarrow left|z^3+1right|&leq R^3+1 \
frac1&geqfrac1R^3+1 \
frac&geqfracR^3+1 textz \
left|fraczz^3+1right|&geqfracRR^3+1 \
endalign
I'm unsure of how to yield the desired inequality. Have I made a mistake somewhere?
EDIT
If I used the inequality $$|z_1+z_2|geq |z_1|-|z_2|$$ I believe this will work. My question is, does this inequality hold for real numbers as well?
complex-analysis proof-verification inequality
asked Aug 3 at 0:53
Bell
560112
1
The triangle $|a|-|b| le |a+b| le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well.
– Doug M
Aug 3 at 1:21
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to show that if $|z|=R>1$, then
$$left|fraczz^3+1right|leqfracRR^3-1$$
I've been playing around with the above inequality and this is where I am at.
beginalign
textConsider left|z^3+1right|&leqleft|z^3right|+1 text(by the triangle inequality)\
&=left|zright|^3+1 \
&=R^3+1 \ \
Rightarrow left|z^3+1right|&leq R^3+1 \
frac1&geqfrac1R^3+1 \
frac&geqfracR^3+1 textz \
left|fraczz^3+1right|&geqfracRR^3+1 \
endalign
I'm unsure of how to yield the desired inequality. Have I made a mistake somewhere?
EDIT
If I used the inequality $$|z_1+z_2|geq |z_1|-|z_2|$$ I believe this will work. My question is, does this inequality hold for real numbers as well?
complex-analysis proof-verification inequality
asked Aug 3 at 0:53
Bell
560112
I am trying to show that if $|z|=R>1$, then
$$left|fraczz^3+1right|leqfracRR^3-1$$
I've been playing around with the above inequality and this is where I am at.
beginalign
textConsider left|z^3+1right|&leqleft|z^3right|+1 text(by the triangle inequality)\
&=left|zright|^3+1 \
&=R^3+1 \ \
Rightarrow left|z^3+1right|&leq R^3+1 \
frac1&geqfrac1R^3+1 \
frac&geqfracR^3+1 textz \
left|fraczz^3+1right|&geqfracRR^3+1 \
endalign
I'm unsure of how to yield the desired inequality. Have I made a mistake somewhere?
EDIT
If I used the inequality $$|z_1+z_2|geq |z_1|-|z_2|$$ I believe this will work. My question is, does this inequality hold for real numbers as well?
complex-analysis proof-verification inequality
asked Aug 3 at 0:53
Bell
560112
asked Aug 3 at 0:53
Bell
560112
asked Aug 3 at 0:53
Bell
560112
asked Aug 3 at 0:53
Bell
560112
560112
1
The triangle $|a|-|b| le |a+b| le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well.
– Doug M
Aug 3 at 1:21
add a comment |Â
1
The triangle $|a|-|b| le |a+b| le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well.
– Doug M
Aug 3 at 1:21
1
1
The triangle $|a|-|b| le |a+b| le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well.
– Doug M
Aug 3 at 1:21
The triangle $|a|-|b| le |a+b| le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well.
– Doug M
Aug 3 at 1:21
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Since $|z|=R$, your inequality reduces to
$$tag1
frac1leqfrac1R^3-1,
$$
which in turn is the same as
$$tag2
R^3-1leq |z^3+1|.
$$
And this is just the reverse triangle inequality:
$$
|z^3+1|geq |z^3|-1=|z|^3-1=R^3-1.
$$
As for your edited question:
$$
|z_1|=|z_1-z_2+z_2|leq|z_1-z_2|+|z_2|.
$$
This gives
$$
|z_1|-|z_2|leq |z_1-z_2|.
$$
Since the roles are interchangeable, we get
$$
|,|z_1|-|z_2|,|leq |z_1-z_2|.
$$
answered Aug 3 at 0:59
Martin Argerami
115k1071164
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
1
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
 |Â
show 1 more comment
up vote
2
down vote
Your inequality is correct, but it's going the wrong way. You want an inequality that ends up with $|z^3+1|$ being greater than something. Try starting with:
$$
|z^3| = |z^3 +1 + (-1)|le |z^3+1| + |-1|
$$
answered Aug 3 at 1:00
grand_chat
17.8k11120
add a comment |Â
up vote
1
down vote
WLOG $z=R(cos t+isin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3cos3tge R^6+1-2R^3= (R^3-1)^2$$
as $cos3tge-1$ and $R^3>1$
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Since $|z|=R$, your inequality reduces to
$$tag1
frac1leqfrac1R^3-1,
$$
which in turn is the same as
$$tag2
R^3-1leq |z^3+1|.
$$
And this is just the reverse triangle inequality:
$$
|z^3+1|geq |z^3|-1=|z|^3-1=R^3-1.
$$
As for your edited question:
$$
|z_1|=|z_1-z_2+z_2|leq|z_1-z_2|+|z_2|.
$$
This gives
$$
|z_1|-|z_2|leq |z_1-z_2|.
$$
Since the roles are interchangeable, we get
$$
|,|z_1|-|z_2|,|leq |z_1-z_2|.
$$
answered Aug 3 at 0:59
Martin Argerami
115k1071164
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
1
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
 |Â
show 1 more comment
up vote
3
down vote
accepted
Since $|z|=R$, your inequality reduces to
$$tag1
frac1leqfrac1R^3-1,
$$
which in turn is the same as
$$tag2
R^3-1leq |z^3+1|.
$$
And this is just the reverse triangle inequality:
$$
|z^3+1|geq |z^3|-1=|z|^3-1=R^3-1.
$$
As for your edited question:
$$
|z_1|=|z_1-z_2+z_2|leq|z_1-z_2|+|z_2|.
$$
This gives
$$
|z_1|-|z_2|leq |z_1-z_2|.
$$
Since the roles are interchangeable, we get
$$
|,|z_1|-|z_2|,|leq |z_1-z_2|.
$$
answered Aug 3 at 0:59
Martin Argerami
115k1071164
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
1
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Since $|z|=R$, your inequality reduces to
$$tag1
frac1leqfrac1R^3-1,
$$
which in turn is the same as
$$tag2
R^3-1leq |z^3+1|.
$$
And this is just the reverse triangle inequality:
$$
|z^3+1|geq |z^3|-1=|z|^3-1=R^3-1.
$$
As for your edited question:
$$
|z_1|=|z_1-z_2+z_2|leq|z_1-z_2|+|z_2|.
$$
This gives
$$
|z_1|-|z_2|leq |z_1-z_2|.
$$
Since the roles are interchangeable, we get
$$
|,|z_1|-|z_2|,|leq |z_1-z_2|.
$$
answered Aug 3 at 0:59
Martin Argerami
115k1071164
Since $|z|=R$, your inequality reduces to
$$tag1
frac1leqfrac1R^3-1,
$$
which in turn is the same as
$$tag2
R^3-1leq |z^3+1|.
$$
And this is just the reverse triangle inequality:
$$
|z^3+1|geq |z^3|-1=|z|^3-1=R^3-1.
$$
As for your edited question:
$$
|z_1|=|z_1-z_2+z_2|leq|z_1-z_2|+|z_2|.
$$
This gives
$$
|z_1|-|z_2|leq |z_1-z_2|.
$$
Since the roles are interchangeable, we get
$$
|,|z_1|-|z_2|,|leq |z_1-z_2|.
$$
answered Aug 3 at 0:59
Martin Argerami
115k1071164
edited Aug 3 at 1:51
answered Aug 3 at 0:59
Martin Argerami
115k1071164
answered Aug 3 at 0:59
Martin Argerami
115k1071164
answered Aug 3 at 0:59
Martin Argerami
115k1071164
115k1071164
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
1
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
 |Â
show 1 more comment
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
1
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
I'm having trouble understanding the reverse triangle inequality (that we call the circle inequality for some reason) step. Doesn't the RTI say $|z_1-z_2|leq left||z|-|z_2|right|$? How did you obtain $|z^3+1|geq|z^3|-1$?
– Bell
Aug 3 at 1:20
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
$z^3=z_1$, $1=z_2$. And no, the inequality you wrote is false most of the time: it implies that if $|z_1|=|z_2|$, then $z_1=z_2$.
– Martin Argerami
Aug 3 at 1:26
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
I think I typed this wrong, I meant $left||z_1|-|z_2|right|leq|z_1-z_2|$. Is this correct?
– Bell
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
Yes, that's exactly what I used and proved in my answer.
– Martin Argerami
Aug 3 at 1:29
1
1
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
Yes, you are right; with $z_2=-1$. You can always drop the "small" absolute value, when dealing with real numbers: if $|a|geq |b|$, then $|a|geq b$. The place where $R>1$ place a role is to go from my inequality $(2)$ to my inequality $(1)$.
– Martin Argerami
Aug 3 at 1:50
 |Â
show 1 more comment
up vote
2
down vote
Your inequality is correct, but it's going the wrong way. You want an inequality that ends up with $|z^3+1|$ being greater than something. Try starting with:
$$
|z^3| = |z^3 +1 + (-1)|le |z^3+1| + |-1|
$$
answered Aug 3 at 1:00
grand_chat
17.8k11120
add a comment |Â
up vote
2
down vote
Your inequality is correct, but it's going the wrong way. You want an inequality that ends up with $|z^3+1|$ being greater than something. Try starting with:
$$
|z^3| = |z^3 +1 + (-1)|le |z^3+1| + |-1|
$$
answered Aug 3 at 1:00
grand_chat
17.8k11120
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your inequality is correct, but it's going the wrong way. You want an inequality that ends up with $|z^3+1|$ being greater than something. Try starting with:
$$
|z^3| = |z^3 +1 + (-1)|le |z^3+1| + |-1|
$$
answered Aug 3 at 1:00
grand_chat
17.8k11120
Your inequality is correct, but it's going the wrong way. You want an inequality that ends up with $|z^3+1|$ being greater than something. Try starting with:
$$
|z^3| = |z^3 +1 + (-1)|le |z^3+1| + |-1|
$$
answered Aug 3 at 1:00
grand_chat
17.8k11120
answered Aug 3 at 1:00
grand_chat
17.8k11120
answered Aug 3 at 1:00
grand_chat
17.8k11120
answered Aug 3 at 1:00
grand_chat
17.8k11120
17.8k11120
add a comment |Â
add a comment |Â
up vote
1
down vote
WLOG $z=R(cos t+isin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3cos3tge R^6+1-2R^3= (R^3-1)^2$$
as $cos3tge-1$ and $R^3>1$
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
add a comment |Â
up vote
1
down vote
WLOG $z=R(cos t+isin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3cos3tge R^6+1-2R^3= (R^3-1)^2$$
as $cos3tge-1$ and $R^3>1$
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
add a comment |Â
up vote
1
down vote
up vote
1
down vote
WLOG $z=R(cos t+isin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3cos3tge R^6+1-2R^3= (R^3-1)^2$$
as $cos3tge-1$ and $R^3>1$
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
WLOG $z=R(cos t+isin t)$ where $t$ is real
$$|z^3+1|^2=R^6+1+2R^3cos3tge R^6+1-2R^3= (R^3-1)^2$$
as $cos3tge-1$ and $R^3>1$
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
answered Aug 3 at 1:15
lab bhattacharjee
214k14152263
214k14152263
add a comment |Â
add a comment |Â
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1
The triangle $|a|-|b| le |a+b| le |a|+|b|$ is all you need. And, yes, it holds for real numbers as well.
– Doug M
Aug 3 at 1:21