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Triangle inequality/complex numbers problem

Hi,

I am faced with a problem. I believe this is true but cannot prove it:

Let z1,z2∈C.z_1, z_2 \in \mathbb{C}.

Suppose ∣z1∣=∣z2∣=1|z_1|=|z_2|=1.

Then ∣z12+z1z2∣≤∣1+z1z2∣|z_1^2+z_1z_2| \leq |1+z_1z_2|.

Any ideas?

Many thanks.
Original post by Zii
Hi,

I am faced with a problem. I believe this is true but cannot prove it:

Let z1,z2∈C.z_1, z_2 \in \mathbb{C}.

Suppose ∣z1∣=∣z2∣=1|z_1|=|z_2|=1.

Then ∣z12+z1z2∣≤∣1+z1z2∣|z_1^2+z_1z_2| \leq |1+z_1z_2|.

Any ideas?

Many thanks.

Your theorem is false, I'm afraid. You can verify that z1=−262137262144+i3669967262144,z2=−131065131072−i1834959131072z_1 = -\frac{262137}{262144} + i \frac{\sqrt{3669967}}{262144}, z_2 = -\frac{131065}{131072} - i \frac{\sqrt{1834959}}{131072} is a counterexample.
Original post by Smaug123
Your theorem is false, I'm afraid. You can verify that z1=−262137262144+i3669967262144,z2=−131065131072−i1834959131072z_1 = -\frac{262137}{262144} + i \frac{\sqrt{3669967}}{262144}, z_2 = -\frac{131065}{131072} - i \frac{\sqrt{1834959}}{131072} is a counterexample.

Please provide another example but with MORE numbers, since this was too easy to plug in to the calculator...
Original post by gagafacea1
Please provide another example but with MORE numbers, since this was too easy to plug in to the calculator...


Sorry, I get a bit whimsical when I'm tired.
Original post by Smaug123
Sorry, I get a bit whimsical when I'm tired.

lol it's fine, it's just that i'm too lazy to test it as well.
Reply 5
Avatar for Zii
Zii
OP
13
In which case the proof that

2≤∣z1+1∣+∣z2+1∣+∣z1z2+1∣2 \leq |z_1+1| + |z_2+1| + |z_1z_2+1|

where

∣z1∣=∣z2∣=1|z_1|=|z_2|=1

is beyond me...
Original post by Zii
In which case the proof that

2≤∣z1+1∣+∣z2+1∣+∣z1z2+1∣2 \leq |z_1+1| + |z_2+1| + |z_1z_2+1|

where

∣z1∣=∣z2∣=1|z_1|=|z_2|=1

is beyond me...

I don't see a nice way to do this. As disgusting as it sounds, casting it into Cartesian coordinates and solving is the best way I've come up with.

Indeed, the bound is tight (there's an easy example to show equality with 2), so we can't afford to lose any information at all in taking the triangle inequality at any stage. In particular, we may only appeal to the triangle inequality once; after that, we can only take the triangle inequality if the conditions for equality are the same as when we first took it.

By the way, this is equivalent to proving ∣cos⁡(u)∣+∣cos⁡(v)∣+∣cos⁡(t−u)∣≥1|\cos(u)| + |\cos(v)| + |\cos(t-u)| \geq 1. That's because 2∣cos⁡(t2)∣=∣1+eit∣2 |\cos(\frac{t}{2})| = |1+e^{i t}|. I don't know if that is helpful.
Reply 7
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Zii
OP
13
Original post by Smaug123
I don't see a nice way to do this. As disgusting as it sounds, casting it into Cartesian coordinates and solving is the best way I've come up with.

Indeed, the bound is tight (there's an easy example to show equality with 2), so we can't afford to lose any information at all in taking the triangle inequality at any stage. In particular, we may only appeal to the triangle inequality once; after that, we can only take the triangle inequality if the conditions for equality are the same as when we first took it.

By the way, this is equivalent to proving ∣cos⁡(u)∣+∣cos⁡(v)∣+∣cos⁡(t−u)∣≥1|\cos(u)| + |\cos(v)| + |\cos(t-u)| \geq 1. That's because 2∣cos⁡(t2)∣=∣1+eit∣2 |\cos(\frac{t}{2})| = |1+e^{i t}|. I don't know if that is helpful.


I get that it should be equivalent to proving
∣cos⁡(u)∣+∣cos⁡(v)∣+∣cos⁡(u+v)∣≥1.|\cos(u)|+|\cos(v)|+|\cos(u+v)| \geq 1.
Not sure where tt has come from in your suggestion.
Original post by Zii
I get that it should be equivalent to proving
∣cos⁡(u)∣+∣cos⁡(v)∣+∣cos⁡(u+v)∣≥1.|\cos(u)|+|\cos(v)|+|\cos(u+v)| \geq 1.
Not sure where tt has come from in your suggestion.

Apologies - entirely my bad. I didn't proofread my answer :P
Reply 9
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Zii
OP
13
Original post by Smaug123
Apologies - entirely my bad. I didn't proofread my answer :P


Your original counterexample doesn't seem to work. Try typing

x=-262137/262144+sqrt(3669967)i/262144, y=-131065/131072-sqrt(1834959)i/131072, abs(x), abs(y), abs(x^2+xy), abs(1+xy)

into WolframAlpha. Then scroll down and click "Approximate form". You will see that

∣x2+xy∣≈1.99992|x^2+xy| \approx 1.99992

whilst

∣1+xy∣≈2.|1+xy| \approx 2.

And I've solved the main problem, namely that if z1,z2∈Cz_1,z_2 \in \mathbb{C} with ∣z1∣=∣z2∣=1,|z_1|=|z_2|=1, then

2≤∣z1+1∣+∣z2+1∣+∣z1z2+1∣2 \leq |z_1+1| + |z_2+1| + |z_1z_2+1|.

How I achieved this:

2=∣1+1∣=∣1+1+z1z2−z1z2+z1−z1∣2 = |1+1| = |1+1+z_1z_2-z_1z_2+z_1-z_1|

=∣z1+1−z1z2−z1+z1z2+1∣≤∣z1+1∣+∣z1z2+z1∣+∣z1z2+1∣= |z_1+1-z_1z_2-z_1+z_1z_2+1| \leq |z_1+1| + |z_1z_2+z_1| + |z_1z_2+1|

=∣z1+1∣+∣z1∣∣z2+1∣+∣z1z2+1∣=∣z1+1∣+∣z2+1∣+∣z1z2+1∣= |z_1+1| + |z_1||z_2+1| + |z_1z_2+1| = |z_1+1| + |z_2+1| + |z_1z_2+1|

as required.
Original post by Zii
Your original counterexample doesn't seem to work. Try typing

x=-262137/262144+sqrt(3669967)i/262144, y=-131065/131072-sqrt(1834959)i/131072, abs(x), abs(y), abs(x^2+xy), abs(1+xy)

into WolframAlpha. Then scroll down and click "Approximate form". You will see that

∣x2+xy∣≈1.99992|x^2+xy| \approx 1.99992

whilst

∣1+xy∣≈2.|1+xy| \approx 2.

And I've solved the main problem, namely that if z1,z2∈Cz_1,z_2 \in \mathbb{C} with ∣z1∣=∣z2∣=1,|z_1|=|z_2|=1, then

2≤∣z1+1∣+∣z2+1∣+∣z1z2+1∣2 \leq |z_1+1| + |z_2+1| + |z_1z_2+1|.

How I achieved this:

2=∣1+1∣=∣1+1+z1z2−z1z2+z1−z1∣2 = |1+1| = |1+1+z_1z_2-z_1z_2+z_1-z_1|

=∣z1+1−z1z2−z1+z1z2+1∣≤∣z1+1∣+∣z1z2+z1∣+∣z1z2+1∣= |z_1+1-z_1z_2-z_1+z_1z_2+1| \leq |z_1+1| + |z_1z_2+z_1| + |z_1z_2+1|

=∣z1+1∣+∣z1∣∣z2+1∣+∣z1z2+1∣=∣z1+1∣+∣z2+1∣+∣z1z2+1∣= |z_1+1| + |z_1||z_2+1| + |z_1z_2+1| = |z_1+1| + |z_2+1| + |z_1z_2+1|

as required.

Good job.
I'm sorry - last night I appear to have fallen asleep some time before I closed my computer. I have no better excuse than that.

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