Mixing
Showing $left|frace^izz^2+1right|leqfrac1^2+1$
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up vote
3
down vote
favorite
I wish to show that if $z$ is real, then
$$left|frace^izz^2+1right|leqfrac1z$$
I have shown this result, although my inequality is the wrong way around.
I considered
beginalign
|z^2+1|&leq |z^2|+|1| text(triangle inequality) \
&=|z|^2+1 \ \
Rightarrow |z^2+1|&leq |z|^2+1 \
frac1&geqfrac1z \
frace^iz&geqfrace^izz \
left|frace^izz^2+1right|&geqfrace^-textIm(z)z \
left|frace^izz^2+1right|&geqfrac1z text(if $z$ is real $Rightarrow$ Im$(z)=0$) \
endalign
Where did I go wrong?
Also, I wonder, would this inequality still hold if $z$ was not real?
proof-verification inequality complex-numbers
edited Aug 3 at 9:45
user 108128
18.7k41544
asked Aug 3 at 9:33
Bell
560112
add a comment |Â
up vote
3
down vote
favorite
I wish to show that if $z$ is real, then
$$left|frace^izz^2+1right|leqfrac1z$$
I have shown this result, although my inequality is the wrong way around.
I considered
beginalign
|z^2+1|&leq |z^2|+|1| text(triangle inequality) \
&=|z|^2+1 \ \
Rightarrow |z^2+1|&leq |z|^2+1 \
frac1&geqfrac1z \
frace^iz&geqfrace^izz \
left|frace^izz^2+1right|&geqfrace^-textIm(z)z \
left|frace^izz^2+1right|&geqfrac1z text(if $z$ is real $Rightarrow$ Im$(z)=0$) \
endalign
Where did I go wrong?
Also, I wonder, would this inequality still hold if $z$ was not real?
proof-verification inequality complex-numbers
edited Aug 3 at 9:45
user 108128
18.7k41544
asked Aug 3 at 9:33
Bell
560112
1
It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold.
– Shashi
Aug 3 at 9:43
I am not sure about the last intermediate step, but the circle $|e^iz| $ always is $1$.
– Nameless
Aug 3 at 9:54
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I wish to show that if $z$ is real, then
$$left|frace^izz^2+1right|leqfrac1z$$
I have shown this result, although my inequality is the wrong way around.
I considered
beginalign
|z^2+1|&leq |z^2|+|1| text(triangle inequality) \
&=|z|^2+1 \ \
Rightarrow |z^2+1|&leq |z|^2+1 \
frac1&geqfrac1z \
frace^iz&geqfrace^izz \
left|frace^izz^2+1right|&geqfrace^-textIm(z)z \
left|frace^izz^2+1right|&geqfrac1z text(if $z$ is real $Rightarrow$ Im$(z)=0$) \
endalign
Where did I go wrong?
Also, I wonder, would this inequality still hold if $z$ was not real?
proof-verification inequality complex-numbers
edited Aug 3 at 9:45
user 108128
18.7k41544
asked Aug 3 at 9:33
Bell
560112
I wish to show that if $z$ is real, then
$$left|frace^izz^2+1right|leqfrac1z$$
I have shown this result, although my inequality is the wrong way around.
I considered
beginalign
|z^2+1|&leq |z^2|+|1| text(triangle inequality) \
&=|z|^2+1 \ \
Rightarrow |z^2+1|&leq |z|^2+1 \
frac1&geqfrac1z \
frace^iz&geqfrace^izz \
left|frace^izz^2+1right|&geqfrace^-textIm(z)z \
left|frace^izz^2+1right|&geqfrac1z text(if $z$ is real $Rightarrow$ Im$(z)=0$) \
endalign
Where did I go wrong?
Also, I wonder, would this inequality still hold if $z$ was not real?
proof-verification inequality complex-numbers
edited Aug 3 at 9:45
user 108128
18.7k41544
asked Aug 3 at 9:33
Bell
560112
edited Aug 3 at 9:45
user 108128
18.7k41544
edited Aug 3 at 9:45
user 108128
18.7k41544
edited Aug 3 at 9:45
user 108128
18.7k41544
18.7k41544
asked Aug 3 at 9:33
Bell
560112
asked Aug 3 at 9:33
Bell
560112
asked Aug 3 at 9:33
Bell
560112
560112
1
It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold.
– Shashi
Aug 3 at 9:43
I am not sure about the last intermediate step, but the circle $|e^iz| $ always is $1$.
– Nameless
Aug 3 at 9:54
add a comment |Â
1
It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold.
– Shashi
Aug 3 at 9:43
I am not sure about the last intermediate step, but the circle $|e^iz| $ always is $1$.
– Nameless
Aug 3 at 9:54
1
1
It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold.
– Shashi
Aug 3 at 9:43
It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold.
– Shashi
Aug 3 at 9:43
I am not sure about the last intermediate step, but the circle $|e^iz| $ always is $1$.
– Nameless
Aug 3 at 9:54
I am not sure about the last intermediate step, but the circle $|e^iz| $ always is $1$.
– Nameless
Aug 3 at 9:54
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
We have
$$left|frace^izz^2+1right|=fracleft= frac1left$$
and
$$0leleft|z^2+1right|= |z|^2+1$$
therefore the result follows.
answered Aug 3 at 9:47
gimusi
63.7k73480
1
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
1
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
1
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
1
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
1
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
 |Â
show 8 more comments
up vote
2
down vote
For real $z$ we have $|z|^2=z^2$ and $|e^iz|$=1, hence
$$left|frace^izz^2+1right|=frac1z.$$
answered Aug 3 at 9:43
Fred
37k1237
add a comment |Â
up vote
2
down vote
If $z$ is real: $|e^iz|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$left|frace^izz^2+1right|=frac1z.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^iz|
=e^-y$ is huge, and so
$$left|frace^izz^2+1right|ggfrac1z.$$
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
add a comment |Â
up vote
1
down vote
$geq $ and $leq$ don't contradict each other! When $z$ is real $|z^2+1|=|z|^2+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^-in$ where $n$ is a large positive integer.
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have
$$left|frace^izz^2+1right|=fracleft= frac1left$$
and
$$0leleft|z^2+1right|= |z|^2+1$$
therefore the result follows.
answered Aug 3 at 9:47
gimusi
63.7k73480
1
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
1
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
1
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
1
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
1
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
 |Â
show 8 more comments
up vote
1
down vote
accepted
We have
$$left|frace^izz^2+1right|=fracleft= frac1left$$
and
$$0leleft|z^2+1right|= |z|^2+1$$
therefore the result follows.
answered Aug 3 at 9:47
gimusi
63.7k73480
1
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
1
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
1
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
1
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
1
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
 |Â
show 8 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have
$$left|frace^izz^2+1right|=fracleft= frac1left$$
and
$$0leleft|z^2+1right|= |z|^2+1$$
therefore the result follows.
answered Aug 3 at 9:47
gimusi
63.7k73480
We have
$$left|frace^izz^2+1right|=fracleft= frac1left$$
and
$$0leleft|z^2+1right|= |z|^2+1$$
therefore the result follows.
answered Aug 3 at 9:47
gimusi
63.7k73480
edited Aug 3 at 9:59
answered Aug 3 at 9:47
gimusi
63.7k73480
answered Aug 3 at 9:47
gimusi
63.7k73480
answered Aug 3 at 9:47
gimusi
63.7k73480
63.7k73480
1
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
1
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
1
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
1
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
1
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
 |Â
show 8 more comments
1
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
1
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
1
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
1
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
1
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
1
1
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
Why doesn't $|e^iz|=1$?
– Bell
Aug 3 at 9:52
1
1
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
@Bell Opsss...yes of course $|e^iz|=1$
– gimusi
Aug 3 at 9:59
1
1
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
@Bell Yes of course bot inequalities are true and indeed $$(age b) land (ale b) iff a=b$$
– gimusi
Aug 3 at 13:32
1
1
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
@Bell Yes exactly!
– gimusi
Aug 3 at 13:40
1
1
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
@Bell You are welcome! It's really a plesure can be useful to you and I also learn a lot dealing with your questions here. Bye
– gimusi
Aug 3 at 13:47
 |Â
show 8 more comments
up vote
2
down vote
For real $z$ we have $|z|^2=z^2$ and $|e^iz|$=1, hence
$$left|frace^izz^2+1right|=frac1z.$$
answered Aug 3 at 9:43
Fred
37k1237
add a comment |Â
up vote
2
down vote
For real $z$ we have $|z|^2=z^2$ and $|e^iz|$=1, hence
$$left|frace^izz^2+1right|=frac1z.$$
answered Aug 3 at 9:43
Fred
37k1237
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For real $z$ we have $|z|^2=z^2$ and $|e^iz|$=1, hence
$$left|frace^izz^2+1right|=frac1z.$$
answered Aug 3 at 9:43
Fred
37k1237
For real $z$ we have $|z|^2=z^2$ and $|e^iz|$=1, hence
$$left|frace^izz^2+1right|=frac1z.$$
answered Aug 3 at 9:43
Fred
37k1237
answered Aug 3 at 9:43
Fred
37k1237
answered Aug 3 at 9:43
Fred
37k1237
answered Aug 3 at 9:43
Fred
37k1237
37k1237
add a comment |Â
add a comment |Â
up vote
2
down vote
If $z$ is real: $|e^iz|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$left|frace^izz^2+1right|=frac1z.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^iz|
=e^-y$ is huge, and so
$$left|frace^izz^2+1right|ggfrac1z.$$
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
add a comment |Â
up vote
2
down vote
If $z$ is real: $|e^iz|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$left|frace^izz^2+1right|=frac1z.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^iz|
=e^-y$ is huge, and so
$$left|frace^izz^2+1right|ggfrac1z.$$
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If $z$ is real: $|e^iz|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$left|frace^izz^2+1right|=frac1z.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^iz|
=e^-y$ is huge, and so
$$left|frace^izz^2+1right|ggfrac1z.$$
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
If $z$ is real: $|e^iz|=1$, $|z^2+1|=z^2+1=|z|^2+1$, so that
$$left|frace^izz^2+1right|=frac1z.$$
If $z=x+iy$ is complex, with $y$ a large negative number, then $|e^iz|
=e^-y$ is huge, and so
$$left|frace^izz^2+1right|ggfrac1z.$$
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 9:45
Lord Shark the Unknown
84.1k950111
84.1k950111
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
add a comment |Â
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
So in terms of the actual inequality in the question, is this possible to obtain? Or is it a strict equality?
– Bell
Aug 3 at 9:53
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
The inequality in the question is true for real $z$, since for real $z$ both sides are equal. It fails for complex $z$; the reverse inequality fails too.
– Lord Shark the Unknown
Aug 3 at 9:55
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
I agree that the $left|frace^izz^2+1right|=frac1z$, but I don't understand why $left|frace^izz^2+1right|leqfrac1z$ for real $z$.
– Bell
Aug 3 at 9:58
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
@Bell If $a=b$ then $ale b$.
– Lord Shark the Unknown
Aug 3 at 10:00
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
I'm a bit embarrassed to say that I don't remember this result. Do you know of a link that may explain it in greater depth? I don't really understand it.
– Bell
Aug 3 at 10:02
add a comment |Â
up vote
1
down vote
$geq $ and $leq$ don't contradict each other! When $z$ is real $|z^2+1|=|z|^2+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^-in$ where $n$ is a large positive integer.
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
1
down vote
$geq $ and $leq$ don't contradict each other! When $z$ is real $|z^2+1|=|z|^2+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^-in$ where $n$ is a large positive integer.
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$geq $ and $leq$ don't contradict each other! When $z$ is real $|z^2+1|=|z|^2+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^-in$ where $n$ is a large positive integer.
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
$geq $ and $leq$ don't contradict each other! When $z$ is real $|z^2+1|=|z|^2+1$. This proves that the stated inequality is actually an equlity for real $z$. To see that the inequality may not hold for complex $z$ take $z=e^-in$ where $n$ is a large positive integer.
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
answered Aug 3 at 9:43
Kavi Rama Murthy
19.2k2829
19.2k2829
add a comment |Â
add a comment |Â
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1
It seems to me that there is an equality. Your consideration is not wrong. For complex $z$ the LHS is unbounded while the RHS is bounded, hence the inequality does not hold.
– Shashi
Aug 3 at 9:43
I am not sure about the last intermediate step, but the circle $|e^iz| $ always is $1$.
– Nameless
Aug 3 at 9:54