Complex Conjugate Inequality Proof

  • #1

Homework Statement


$$
\left | \frac{z}{\left | z \right |} + \frac{w}{\left | w \right |} \right |\left ( \left | z \right | +\left | w \right | \right )\leq 2\left | z+w \right |
$$
Where z and w are complex numbers not equal to zero.
2.$$\frac{z}{\left | z \right | ^{2}}=\frac{1}{\bar{z}}$$

The Attempt at a Solution


EDIT: INCORRECT WORK[/B]
Alright, the following was my approach. I first tried to split up the problem due to symmetry.
$$
\left | \frac{z}{\left | z \right |} + \frac{w}{\left | w \right |} \right | \left | z \right | \leq \left | z+w \right |
$$
$$
\left | \frac{1}{\bar{z}} + \frac{1}{\bar{w}} \right | \left | z \right | \leq \left | z+w \right |

$$
$$
\left | \frac{\bar{z}+\bar{w}}{\bar{z}\bar{w}} \right | \left | z \right | \leq \left | z+w \right |
$$
And I was left with the following
$$
\left | \bar{z}+\bar{w} \right | \left | z \right | \leq \left | z+w \right | |
\bar{z}\bar{w}|
$$

I feel like I'm either close, or I've worked the problem entirely the wrong way including conjugates.
 
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Answers and Replies

  • #2
Charles Link
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I don't think you can split the problem by symmetry in such a manner. I would reconsider that step. If the one term is equal to a where a<1, it means its symmetric pair just needs to be less than 2-a. If you could prove the one term is less than one, it would guarantee the sum is less than 2, but that is probably not the case. Most likely, the single term is not necessarily less than one.
 
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  • #3
I don't think you can split the problem by symmetry in such a manner. I would reconsider that step. If the one term is equal to a where a<1, it means its symmetric pair just needs to be less than 2-a. If you could prove the one term is less than one, it would guarantee the sum is less than 2, but that is probably not the case. Most likely, the single term is not necessarily less than one.
Hmm, yes I see what you mean. I'm really running out of ideas on how to operate. I guess I could try expanding the binomials, but that looks like a long and arduous task (one I am willing to do).
 
  • #4
Charles Link
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Hmm, yes I see what you mean. I'm really running out of ideas on how to operate. I guess I could try expanding the binomials, but that looks like a long and arduous task (one I am willing to do).
Suggest you square both sides because |w+z| gives you a square root. Incidentally in your second line in post 1, that should be ## |z|^2 ## in the denominator.
 
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  • #5
Suggest you square both sides because |w+z| gives you a square root. Incidentally in your second line in post 1, that should be ## |z|^2 ## in the denominator.
Yup, looks like I goofed. It does not change anything though does it? Symmetry was no good...
 
  • #6
Charles Link
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Yup, looks like I goofed. It does not change anything though does it? Symmetry was no good...
It looks like a complete expansion with squaring is your best approach. I expect terms will cancel on both sides. It is possible you will need to square a second time after that, but I do think this one will not be as difficult as it looks.
 
  • #7
Ray Vickson
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Homework Statement


$$
\left | \frac{z}{\left | z \right |} + \frac{w}{\left | w \right |} \right |\left ( \left | z \right | +\left | w \right | \right )\leq 2\left | z+w \right |
$$
Where z and w are complex numbers not equal to zero.
2.$$\frac{z}{\left | z \right |}=\frac{1}{\bar{z}}$$

The Attempt at a Solution


Alright, the following was my approach. I first tried to split up the problem due to symmetry.
$$
\left | \frac{z}{\left | z \right |} + \frac{w}{\left | w \right |} \right | \left | z \right | \leq \left | z+w \right |
$$
$$
\left | \frac{1}{\bar{z}} + \frac{1}{\bar{w}} \right | \left | z \right | \leq \left | z+w \right |

$$
$$
\left | \frac{\bar{z}+\bar{w}}{\bar{z}\bar{w}} \right | \left | z \right | \leq \left | z+w \right |
$$
And I was left with the following
$$
\left | \bar{z}+\bar{w} \right | \left | z \right | \leq \left | z+w \right | |
\bar{z}\bar{w}|
$$

I feel like I'm either close, or I've worked the problem entirely the wrong way including conjugates.

If you write ##z = r_1 e^{i \theta_1}## and ##w = r_2 e^{i \theta_2}## the left and right-hand sides of your proposed inequality become quite simple functions of ##r_1, r_2, \theta_1, \theta_2##. You can use trigonometric identities and a bit of algebra to get a relatively simple inequality to verify.
 
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  • #8
If you write ##z = r_1 e^{i \theta_1}## and ##w = r_2 e^{i \theta_2}## the left and right-hand sides of your proposed inequality become quite simple functions of ##r_1, r_2, \theta_1, \theta_2##. You can use trigonometric identities and a bit of algebra to get a relatively simple inequality to verify.

By proposed inequality you mean the initial inequality we are trying to prove, right? The work leading up to the bottom bit is incorrect.
 
  • #9
Ray Vickson
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By proposed inequality you mean the initial inequality we are trying to prove, right? The work leading up to the bottom bit is incorrect.

Yes, the proposed inequality is the one you are trying to prove.

When you say the work leading up to the bottom bit is incorrect, whose bottom bit are you referring to? What message number are you citing?

By using the polar representation I was able to verify your proposed inequality using only elementary algebra, trigonometry and (at one point) elementary calculus. It takes a bit of work, but it really does go through. However, I would not be comfortable with saying much more at this stage, due to PF rules, etc.
 
  • #10
Yes, the proposed inequality is the one you are trying to prove.

When you say the work leading up to the bottom bit is incorrect, whose bottom bit are you referring to? What message number are you citing?

By using the polar representation I was able to verify your proposed inequality using only elementary algebra, trigonometry and (at one point) elementary calculus. It takes a bit of work, but it really does go through. However, I would not be comfortable with saying much more at this stage, due to PF rules, etc.
So I went ahead and did the problem yesterday, I believe it ended well. Didn't prove one part though, when I divide both sides by double sines and cosines, I don't check if it's less than 0. https://drive.google.com/open?id=0BxlGAHNyMIneeUpXWEluZWF0TEE
 
  • #12
Charles Link
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(Explanations are in spanish, though my guess is you guys will figure it out)
Your ## cos(\theta)cos(\phi)-sin(\theta)sin(\phi) = cos(\theta +\phi) ## so it necessarily has absolute value less than ((edited:) or equal to )1 and won't change the sign of your inequality.
 
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  • #13
Your ## cos(\theta)cos(\phi)-sin(\theta)sin(\phi) = cos(\theta +\phi) ## so it necessarily has absolute value less than 1 and won't change the sign of your inequality.
True, I think I'll add it to my proof. Thanks Charles (Credit where due)
 
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  • #14
Ray Vickson
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So I went ahead and did the problem yesterday, I believe it ended well. Didn't prove one part though, when I divide both sides by double sines and cosines, I don't check if it's less than 0. https://drive.google.com/open?id=0BxlGAHNyMIneeUpXWEluZWF0TEE
Here is what I did: if ##w = re^{i \lambda}## and ##z = k r e^{i \phi}## (##r,k > 0##) then
$$L \equiv \left| \frac{w}{|w|} + \frac{z}{|z|} \right| (|w|+|z|) = r(1+k) |e^{i \lambda} + e^{ i \phi}|$$
and
$$R \equiv 2 |w+z| = 2 |r e^{i \lambda} + rk e^{i \phi}|$$.
Without loss of generality we can assume ##\lambda = 0##; if not, just use the fact that
$$ L = |e^{-i\lambda}|L= r(1+k) |e^{-i \lambda}(e^{i \lambda} + e^{ i \phi})| $$
and
$$ R = |e^{- i \lambda}| R = 2 |e^{-i \lambda}(r e^{i \lambda} + rk e^{i \phi})|$$
use the new variables ##w = r## and ##z = k r e^{i \theta}##, where ##\theta = \phi - \lambda##.

We have
$$L = r (1+k) \sqrt{2(1+c)},\; R = 2 r \sqrt{k^2+1+2 k c},$$
where ##c = \cos(\theta)##. Thus,
$$ \begin{array}{rcl}R-L &=& r [ 2 \sqrt{1+k^2 + 2kc} - (1+k) \sqrt{2(1+c)}] \\
&=& r \sqrt{1+c}\, f(k,c),
\end{array}$$
where
$$f(k,c) = 2 \sqrt{ 2k + (k-1)^2/(1+c)} - \sqrt{2} (k+1) $$
(because ##1+k^2+2kc = (k-1)^2 + 2k(1+c)##).

For any fixed ##k > 0## the function ##f(k,c)## is strictly decreasing in ##c## on the interval ##-1 < c \leq 1##, so for ##-1 < c < 1## we have ##f(k,c) > f(k,1) = 0##. Thus, ##R-L > 0## for ##-1 \leq c < 1##, and ##R=L## for ##c = 1 \; (\theta = 0)##.
 
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