Mixing
Complex analysis proof triangle inequality:
$begingroup$
Given: $|z+w|^2=|z|^2+|w|^2+2Re(zbar w)$
Prove:$|z+w|leq |z|+|w|$
Work done so far:
Let $z=x+iy$ and $w=a+bi$, then:
$$|x+iy+a+ib|=|z+w|=sqrt{(x+a)^2+(y+b)^2}$$
$$sqrt{x^2+y^2}+sqrt{a^2+b^2}=|z|+|w|$$
Squaring it I get,
$$x^2+y^2+2|z||w|+a^2+y^2$$
After this I am lost, any idea how to proceed or if I am doing it wrong, what is the right way.
Any help is appreciated!
complex-analysis proof-writing
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
$endgroup$
add a comment |
$begingroup$
Given: $|z+w|^2=|z|^2+|w|^2+2Re(zbar w)$
Prove:$|z+w|leq |z|+|w|$
Work done so far:
Let $z=x+iy$ and $w=a+bi$, then:
$$|x+iy+a+ib|=|z+w|=sqrt{(x+a)^2+(y+b)^2}$$
$$sqrt{x^2+y^2}+sqrt{a^2+b^2}=|z|+|w|$$
Squaring it I get,
$$x^2+y^2+2|z||w|+a^2+y^2$$
After this I am lost, any idea how to proceed or if I am doing it wrong, what is the right way.
Any help is appreciated!
complex-analysis proof-writing
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
$endgroup$
$begingroup$
Your premise is incorrect... it's not even true if $z = w = 1$.
$endgroup$
– T. Bongers
Jan 24 at 23:04
$begingroup$
Sorry, Made a slight typo. I will fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
@T.Bongers fixed!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
add a comment |
$begingroup$
Given: $|z+w|^2=|z|^2+|w|^2+2Re(zbar w)$
Prove:$|z+w|leq |z|+|w|$
Work done so far:
Let $z=x+iy$ and $w=a+bi$, then:
$$|x+iy+a+ib|=|z+w|=sqrt{(x+a)^2+(y+b)^2}$$
$$sqrt{x^2+y^2}+sqrt{a^2+b^2}=|z|+|w|$$
Squaring it I get,
$$x^2+y^2+2|z||w|+a^2+y^2$$
After this I am lost, any idea how to proceed or if I am doing it wrong, what is the right way.
Any help is appreciated!
complex-analysis proof-writing
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
$endgroup$
Given: $|z+w|^2=|z|^2+|w|^2+2Re(zbar w)$
Prove:$|z+w|leq |z|+|w|$
Work done so far:
Let $z=x+iy$ and $w=a+bi$, then:
$$|x+iy+a+ib|=|z+w|=sqrt{(x+a)^2+(y+b)^2}$$
$$sqrt{x^2+y^2}+sqrt{a^2+b^2}=|z|+|w|$$
Squaring it I get,
$$x^2+y^2+2|z||w|+a^2+y^2$$
After this I am lost, any idea how to proceed or if I am doing it wrong, what is the right way.
Any help is appreciated!
complex-analysis proof-writing
complex-analysis proof-writing
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
edited Jan 24 at 23:05
Bertrand Wittgenstein's Ghost
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
asked Jan 24 at 23:02
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
493215
493215
$begingroup$
Your premise is incorrect... it's not even true if $z = w = 1$.
$endgroup$
– T. Bongers
Jan 24 at 23:04
$begingroup$
Sorry, Made a slight typo. I will fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
@T.Bongers fixed!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
add a comment |
$begingroup$
Your premise is incorrect... it's not even true if $z = w = 1$.
$endgroup$
– T. Bongers
Jan 24 at 23:04
$begingroup$
Sorry, Made a slight typo. I will fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
@T.Bongers fixed!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
Your premise is incorrect... it's not even true if $z = w = 1$.
$endgroup$
– T. Bongers
Jan 24 at 23:04
$begingroup$
Your premise is incorrect... it's not even true if $z = w = 1$.
$endgroup$
– T. Bongers
Jan 24 at 23:04
$begingroup$
Sorry, Made a slight typo. I will fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
Sorry, Made a slight typo. I will fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
@T.Bongers fixed!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
@T.Bongers fixed!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You want to show that
$$tag1
sqrt{(x+a)^2+(y+b)^2}leq sqrt{a^2+b^2}+sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$tag2
{(x+a)^2+(y+b)^2}leq a^2+b^2+x^2+y^2+2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
which reduces to
$$tag3
2ax+2ybleq 2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
and squaring again this is (after cancelling terms),
$$tag4
0leq 4a^2y^2+4b^2x^2.
$$
is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get
$$
(2ax+2by)^2leq 4(a^2+b^2)(x^2+y^2).
$$
Taking square root (everything is positive), we get
$$
2ax+2byleq|2ax+2by|leq2sqrt{a^2+b^2},sqrt{x^2+y^2}.
$$
Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
1
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
1
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
1
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
|
show 1 more comment
$begingroup$
HINT: Please do not write out real and imaginary parts. Just use the inequality you were given. Compare $|z+w|^2$ and $(|z|+|w|)^2$, and recall what you know (or can prove) if $a,bge 0$ and $a^2le b^2$.
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
$begingroup$
Can you please expand on it, thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
1
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
$begingroup$
got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
$begingroup$
Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
add a comment |
$begingroup$
You just need to show that
$$
operatorname{Re}(zbar{w})le |z||w| tag{*}
$$
because if this is true, then
$$
|z+w|^2=|z|^2+|w|^2+2operatorname{Re}(zbar{w})le
|z|^2+|w|^2+2|z||w|=(|z|+|w|)^2
$$
Now (*) is obvious if $operatorname{Re}(zbar{w})<0$. Suppose it is $ge0$. Then, with your notation, you have to prove that
$$
(xa+yb)lesqrt{x^2+y^2}sqrt{a^2+b^2}
$$
Square both sides and conclude.
Alternatively, observe that $lvertoperatorname{Re}{z}rvertlelvert zrvert$; this will lead to a shorter proof.
answered Jan 24 at 23:11
egregegreg
184k1486205
$endgroup$
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
1
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
1
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
1
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You want to show that
$$tag1
sqrt{(x+a)^2+(y+b)^2}leq sqrt{a^2+b^2}+sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$tag2
{(x+a)^2+(y+b)^2}leq a^2+b^2+x^2+y^2+2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
which reduces to
$$tag3
2ax+2ybleq 2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
and squaring again this is (after cancelling terms),
$$tag4
0leq 4a^2y^2+4b^2x^2.
$$
is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get
$$
(2ax+2by)^2leq 4(a^2+b^2)(x^2+y^2).
$$
Taking square root (everything is positive), we get
$$
2ax+2byleq|2ax+2by|leq2sqrt{a^2+b^2},sqrt{x^2+y^2}.
$$
Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
1
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
1
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
1
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
|
show 1 more comment
$begingroup$
You want to show that
$$tag1
sqrt{(x+a)^2+(y+b)^2}leq sqrt{a^2+b^2}+sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$tag2
{(x+a)^2+(y+b)^2}leq a^2+b^2+x^2+y^2+2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
which reduces to
$$tag3
2ax+2ybleq 2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
and squaring again this is (after cancelling terms),
$$tag4
0leq 4a^2y^2+4b^2x^2.
$$
is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get
$$
(2ax+2by)^2leq 4(a^2+b^2)(x^2+y^2).
$$
Taking square root (everything is positive), we get
$$
2ax+2byleq|2ax+2by|leq2sqrt{a^2+b^2},sqrt{x^2+y^2}.
$$
Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
1
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
1
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
1
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
|
show 1 more comment
$begingroup$
You want to show that
$$tag1
sqrt{(x+a)^2+(y+b)^2}leq sqrt{a^2+b^2}+sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$tag2
{(x+a)^2+(y+b)^2}leq a^2+b^2+x^2+y^2+2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
which reduces to
$$tag3
2ax+2ybleq 2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
and squaring again this is (after cancelling terms),
$$tag4
0leq 4a^2y^2+4b^2x^2.
$$
is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get
$$
(2ax+2by)^2leq 4(a^2+b^2)(x^2+y^2).
$$
Taking square root (everything is positive), we get
$$
2ax+2byleq|2ax+2by|leq2sqrt{a^2+b^2},sqrt{x^2+y^2}.
$$
Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
You want to show that
$$tag1
sqrt{(x+a)^2+(y+b)^2}leq sqrt{a^2+b^2}+sqrt{x^2+y^2}.
$$
If you look at the squares, that would be
$$tag2
{(x+a)^2+(y+b)^2}leq a^2+b^2+x^2+y^2+2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
which reduces to
$$tag3
2ax+2ybleq 2sqrt{a^2+b^2},sqrt{x^2+y^2},
$$
and squaring again this is (after cancelling terms),
$$tag4
0leq 4a^2y^2+4b^2x^2.
$$
is obviously true. So, if we start from $(4)$, we add $4a^2x^2+4b^2y^2$ to both sides, to get
$$
(2ax+2by)^2leq 4(a^2+b^2)(x^2+y^2).
$$
Taking square root (everything is positive), we get
$$
2ax+2byleq|2ax+2by|leq2sqrt{a^2+b^2},sqrt{x^2+y^2}.
$$
Now add $a^2+b^2+x^2+y^2$ to both sides, to get $(2)$, and taking square roots again we get $(1)$.
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
edited Jan 25 at 2:03
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 23:13
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
1
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
1
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
1
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
|
show 1 more comment
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
1
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
1
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
1
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
$begingroup$
Can you please fill in the details, I am having a hard time on this question. Thank you.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:40
1
1
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
What details? You don't know how to expand a square?
$endgroup$
– Martin Argerami
Jan 25 at 1:41
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
$begingroup$
As in, how are they related, I can not see the connection. Thanks!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:51
1
1
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
$begingroup$
You are right, I'm sorry. I typed things the wrong way. I'll edit soon.
$endgroup$
– Martin Argerami
Jan 25 at 1:58
1
1
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
$begingroup$
Edited. My apologies.
$endgroup$
– Martin Argerami
Jan 25 at 2:04
|
show 1 more comment
$begingroup$
HINT: Please do not write out real and imaginary parts. Just use the inequality you were given. Compare $|z+w|^2$ and $(|z|+|w|)^2$, and recall what you know (or can prove) if $a,bge 0$ and $a^2le b^2$.
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
$begingroup$
Can you please expand on it, thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
1
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
$begingroup$
got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
$begingroup$
Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
add a comment |
$begingroup$
HINT: Please do not write out real and imaginary parts. Just use the inequality you were given. Compare $|z+w|^2$ and $(|z|+|w|)^2$, and recall what you know (or can prove) if $a,bge 0$ and $a^2le b^2$.
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
$begingroup$
Can you please expand on it, thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
1
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
$begingroup$
got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
$begingroup$
Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
add a comment |
$begingroup$
HINT: Please do not write out real and imaginary parts. Just use the inequality you were given. Compare $|z+w|^2$ and $(|z|+|w|)^2$, and recall what you know (or can prove) if $a,bge 0$ and $a^2le b^2$.
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
$endgroup$
HINT: Please do not write out real and imaginary parts. Just use the inequality you were given. Compare $|z+w|^2$ and $(|z|+|w|)^2$, and recall what you know (or can prove) if $a,bge 0$ and $a^2le b^2$.
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
answered Jan 24 at 23:12
Ted ShifrinTed Shifrin
64.3k44692
64.3k44692
$begingroup$
Can you please expand on it, thanks.
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– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
1
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
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got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
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Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
add a comment |
$begingroup$
Can you please expand on it, thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
1
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
$begingroup$
got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
$begingroup$
Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
$begingroup$
Can you please expand on it, thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
$begingroup$
Can you please expand on it, thanks.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:39
1
1
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
$begingroup$
@egreg already wrote out the argument. If you show $|z+w|^2le (|z|+|w|)^2$, then it follows that $|z+w|le |z|+|w|$. (That's the jist of my final sentence. Why?)
$endgroup$
– Ted Shifrin
Jan 25 at 2:02
$begingroup$
got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
$begingroup$
got it, thank you!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 4:12
$begingroup$
Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
$begingroup$
Thank you for the answer!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:11
add a comment |
$begingroup$
You just need to show that
$$
operatorname{Re}(zbar{w})le |z||w| tag{*}
$$
because if this is true, then
$$
|z+w|^2=|z|^2+|w|^2+2operatorname{Re}(zbar{w})le
|z|^2+|w|^2+2|z||w|=(|z|+|w|)^2
$$
Now (*) is obvious if $operatorname{Re}(zbar{w})<0$. Suppose it is $ge0$. Then, with your notation, you have to prove that
$$
(xa+yb)lesqrt{x^2+y^2}sqrt{a^2+b^2}
$$
Square both sides and conclude.
Alternatively, observe that $lvertoperatorname{Re}{z}rvertlelvert zrvert$; this will lead to a shorter proof.
answered Jan 24 at 23:11
egregegreg
184k1486205
$endgroup$
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
1
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
1
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
1
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
add a comment |
$begingroup$
You just need to show that
$$
operatorname{Re}(zbar{w})le |z||w| tag{*}
$$
because if this is true, then
$$
|z+w|^2=|z|^2+|w|^2+2operatorname{Re}(zbar{w})le
|z|^2+|w|^2+2|z||w|=(|z|+|w|)^2
$$
Now (*) is obvious if $operatorname{Re}(zbar{w})<0$. Suppose it is $ge0$. Then, with your notation, you have to prove that
$$
(xa+yb)lesqrt{x^2+y^2}sqrt{a^2+b^2}
$$
Square both sides and conclude.
Alternatively, observe that $lvertoperatorname{Re}{z}rvertlelvert zrvert$; this will lead to a shorter proof.
answered Jan 24 at 23:11
egregegreg
184k1486205
$endgroup$
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
1
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
1
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
1
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
add a comment |
$begingroup$
You just need to show that
$$
operatorname{Re}(zbar{w})le |z||w| tag{*}
$$
because if this is true, then
$$
|z+w|^2=|z|^2+|w|^2+2operatorname{Re}(zbar{w})le
|z|^2+|w|^2+2|z||w|=(|z|+|w|)^2
$$
Now (*) is obvious if $operatorname{Re}(zbar{w})<0$. Suppose it is $ge0$. Then, with your notation, you have to prove that
$$
(xa+yb)lesqrt{x^2+y^2}sqrt{a^2+b^2}
$$
Square both sides and conclude.
Alternatively, observe that $lvertoperatorname{Re}{z}rvertlelvert zrvert$; this will lead to a shorter proof.
answered Jan 24 at 23:11
egregegreg
184k1486205
$endgroup$
You just need to show that
$$
operatorname{Re}(zbar{w})le |z||w| tag{*}
$$
because if this is true, then
$$
|z+w|^2=|z|^2+|w|^2+2operatorname{Re}(zbar{w})le
|z|^2+|w|^2+2|z||w|=(|z|+|w|)^2
$$
Now (*) is obvious if $operatorname{Re}(zbar{w})<0$. Suppose it is $ge0$. Then, with your notation, you have to prove that
$$
(xa+yb)lesqrt{x^2+y^2}sqrt{a^2+b^2}
$$
Square both sides and conclude.
Alternatively, observe that $lvertoperatorname{Re}{z}rvertlelvert zrvert$; this will lead to a shorter proof.
answered Jan 24 at 23:11
egregegreg
184k1486205
edited Jan 25 at 8:25
answered Jan 24 at 23:11
egregegreg
184k1486205
answered Jan 24 at 23:11
egregegreg
184k1486205
answered Jan 24 at 23:11
egregegreg
184k1486205
184k1486205
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
1
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
1
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
1
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
add a comment |
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
1
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
1
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
1
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
$begingroup$
How should I go about proving the last statement! I can not see how that immediately follows from the premises I have.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 1:52
1
1
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
$begingroup$
You don't need to resort to real and imaginary parts. $|text{Re}(v)|le |v|$ for any complex number, so $text{Re}(zbar w)le |zbar w| = |z||bar w| = |z||w|$.
$endgroup$
– Ted Shifrin
Jan 25 at 2:01
1
1
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
$begingroup$
@BertrandWittgenstein'sGhost Square and simplify: you'll see something.
$endgroup$
– egreg
Jan 25 at 7:51
1
1
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
@TedShifrin Yes, but I'm not sure the OP would believe to that.
$endgroup$
– egreg
Jan 25 at 7:52
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
$begingroup$
Thank you for the answer, I wish I could accept two answers. This helped me immensely.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 25 at 8:12
add a comment |
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$begingroup$
Your premise is incorrect... it's not even true if $z = w = 1$.
$endgroup$
– T. Bongers
Jan 24 at 23:04
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Sorry, Made a slight typo. I will fix it.
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05
$begingroup$
@T.Bongers fixed!
$endgroup$
– Bertrand Wittgenstein's Ghost
Jan 24 at 23:05