Complex inequality $|a+b|le |a|+|b|$
$begingroup$
My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.
How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.
If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.
It is on page 9 of Ahlfors' Complex Analysis.
complex-analysis proof-verification complex-numbers proof-explanation
asked Dec 31 '18 at 21:26
user398843user398843
718316
$endgroup$
add a comment |
$begingroup$
My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.
How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.
If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.
It is on page 9 of Ahlfors' Complex Analysis.
complex-analysis proof-verification complex-numbers proof-explanation
asked Dec 31 '18 at 21:26
user398843user398843
718316
$endgroup$
1
$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33
$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53
$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59
add a comment |
$begingroup$
My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.
How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.
If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.
It is on page 9 of Ahlfors' Complex Analysis.
complex-analysis proof-verification complex-numbers proof-explanation
asked Dec 31 '18 at 21:26
user398843user398843
718316
$endgroup$
My textbook says if $a$ and $b$ are two complex numbers, then $$|a+b|le |a|+|b|,$$ and the equality holds if and only if $abar{b} ge 0$.
How can we say the equality holds if and only if $abar b ge 0$? I think $abar b$ is a complex number and complex numbers do not have order.
If we square both sides and cancel some terms, then we can see that the equality holds if Re$(abar b) = |a||b|$.
It is on page 9 of Ahlfors' Complex Analysis.
complex-analysis proof-verification complex-numbers proof-explanation
complex-analysis proof-verification complex-numbers proof-explanation
asked Dec 31 '18 at 21:26
user398843user398843
718316
asked Dec 31 '18 at 21:26
user398843user398843
718316
asked Dec 31 '18 at 21:26
user398843user398843
718316
asked Dec 31 '18 at 21:26
user398843user398843
718316
asked Dec 31 '18 at 21:26
user398843user398843
718316
718316
1
$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33
$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53
$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59
add a comment |
1
$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33
$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53
$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59
1
1
$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33
$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33
$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53
$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53
$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59
$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that
$$
operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
$$
if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
$endgroup$
1
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
add a comment |
$begingroup$
The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
abar{b}+bar{a}ble2|a|,|b| tag{*}
$$
This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).
Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(abar{b}-bar{a}b)^2le0
$$
which is true, because $abar{b}-bar{a}b$ is purely imaginary.
A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
$endgroup$
add a comment |
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3 Answers
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active
oldest
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3 Answers
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active
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$begingroup$
Note that
$$
operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
$$
if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
$endgroup$
1
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
add a comment |
$begingroup$
Note that
$$
operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
$$
if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
$endgroup$
1
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
add a comment |
$begingroup$
Note that
$$
operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
$$
if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
$endgroup$
Note that
$$
operatorname{Re}(abar b)=|a||b|=|ab|=|abar b|
$$
if and only if $operatorname{Im}(abar b)=0$ and $abar bge 0$, so $abar b$ has to be real.
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
edited Dec 31 '18 at 21:34
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
answered Dec 31 '18 at 21:31
A.Γ.A.Γ.
22.9k32656
22.9k32656
1
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
add a comment |
1
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
1
1
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
If $a=1$, $b=-1$, then $operatorname{Im}(abar b)=0$, but $operatorname{Re}(abar b)ne|abar b|$.
$endgroup$
– Martin Argerami
Dec 31 '18 at 21:33
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
$begingroup$
@MartinArgerami That's true, thanks.
$endgroup$
– A.Γ.
Dec 31 '18 at 21:36
add a comment |
$begingroup$
The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
abar{b}+bar{a}ble2|a|,|b| tag{*}
$$
This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).
Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(abar{b}-bar{a}b)^2le0
$$
which is true, because $abar{b}-bar{a}b$ is purely imaginary.
A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
abar{b}+bar{a}ble2|a|,|b| tag{*}
$$
This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).
Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(abar{b}-bar{a}b)^2le0
$$
which is true, because $abar{b}-bar{a}b$ is purely imaginary.
A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
$endgroup$
add a comment |
$begingroup$
The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
abar{b}+bar{a}ble2|a|,|b| tag{*}
$$
This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).
Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(abar{b}-bar{a}b)^2le0
$$
which is true, because $abar{b}-bar{a}b$ is purely imaginary.
A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
$endgroup$
The inequality holds if and only if $|a+b|^2le(|a|+|b|)^2$ holds, which becomes
$$
|a|^2+abar{b}+bar{a}b+|b|^2le |a|^2+2|a|,|b|+|b|^2
$$
We can cancel real terms from both sides and still get an equivalent inequality:
$$
abar{b}+bar{a}ble2|a|,|b| tag{*}
$$
This surely holds and is strict if $abar{b}+bar{a}b<0$ (following Ahlfors's convention that $c>0$ means that $c$ is real and positive).
Let's suppose $abar{b}+bar{a}bge0$. Then squaring is allowed and yields an equivalent inequality:
$$
a^2bar{b}^2+2abar{a}bbar{b}+bar{a}^2b^2le 4abar{a}bbar{b}
$$
that is, moving the (real) right-hand side to the left
$$
(abar{b}-bar{a}b)^2le0
$$
which is true, because $abar{b}-bar{a}b$ is purely imaginary.
A necessary condition for having an equality is that $abar{b}=bar{a}b$, that is $abar{b}$ is real. Together with (*) we obtain $abar{b}=2|a|,|b|ge0$.
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
answered Dec 31 '18 at 22:07
egregegreg
186k1486209
186k1486209
add a comment |
add a comment |
$begingroup$
Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
$endgroup$
add a comment |
$begingroup$
Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
$endgroup$
add a comment |
$begingroup$
Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
$endgroup$
Think of $a$ and $b$ as being planar vectors. Draw the parallelogram and apply the parallelogram law. You will see the geometric meaning of this immediately.
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
answered Dec 31 '18 at 23:52
ncmathsadistncmathsadist
43.2k261103
43.2k261103
add a comment |
add a comment |
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1
$begingroup$
What they mean with "$abar bgeqslant0$" is that $abar b$ should be a nonnegative real number. Yes, this is sloppy writing.
$endgroup$
– Did
Dec 31 '18 at 21:33
$begingroup$
See the parenthesis in Ahlfors immediately after this point: it is convenient to let $c>0$ indicate that $c$ is real and positive.
$endgroup$
– GEdgar
Dec 31 '18 at 21:53
$begingroup$
We can show $abar b$ is nonnegative real number by expanding $abar b$ and notice that its imaginary part is zero since we have the relation Re$(abar b) = |a||b|$.
$endgroup$
– user398843
Dec 31 '18 at 21:59