Proving Inequality for Complex Numbers with Absolute Value Constraints

[Brunno]

Hi fellows,

Homework Statement

Prove that:


\sqrt{\frac{7}{2}}\leq|z+1|+|1-z+z²|\leq\sqrt{\frac{7}{6}}

for all complex numbers with |z|=1.

Homework Equations

The Attempt at a Solution

I've tried something like this:

Starting by the following property:

-|z|\leqRe(z)\leq|z|

but i could'nt get anywhere.


Thanks in advance.

 

[Brunno]

Hellllllp! Please!:)

 

[Char. Limit]

Have you tried squaring both sides of the inequality?

 

[Brunno]

Yes:

7/2<= |t²+2tu+u²|<=63/6

But from here i can't go anywhere.I don't know from where to start to prove it.:(

where t=(z+1) and u=(1-z+z²)

 

[tiny-tim]

Hi Brunno!

I can't see a neat way of doing it

You could try putting z = cosθ + isinθ, z² = cos2θ + isin2θ.

Or maybe use 1 - z + z² = (1 + z³)/(1 + z).

(I haven't tried either of them)

 

[Brunno]

Hi,
I tried the second one but still doesn't seem to make it easier.How to solve it by the non neat way?

 

[Mark44]

\sqrt{\frac{7}{2}}\leq|z+1|+|1-z+z²|\leq\sqrt{\frac{7}{6}}

for all complex numbers with |z|=1.

This is a relatively minor point, but the direction of your inequalities is going the wrong way. For one thing you are saying that sqrt(7/2) <= sqrt(7/6), which isn't true.

Also, your inequality will look better if you use just a single pair of tex and /tex tags.

Here's your inequality in cleaned up form, with the correction noted above.

\sqrt{\frac{7}{6}} \leq |z+1|+|1-z+z^2| \leq \sqrt{\frac{7}{2}}

Click the inequality to see what my LaTeX script looks like.

 

[Brunno]

Mark44,thanks.I didn't notice it.Yes it's not sqrt(7/6) but actually 3sqrt(7/6).

 

[Brunno]

Anyone willing to help me on this one?;(

 

[Char. Limit]

Yes:

7/2<= |t²+2tu+u²|<=63/6

But from here i can't go anywhere.I don't know from where to start to prove it.:(

where t=(z+1) and u=(1-z+z²)

That "t²+2tu+u²" is better written in terms of z as...

(z²+2)²

That might also make it easier.

 

[Brunno]

So it should be something like:

7/2<=(z²+2)²<=63/6

I still can't see any further...

 

[Brunno]

HeLLLLP People!:)

 

[I like Serena]

HeLLLLP People!:)

Well, there's a few things wrong with the above.

For starters, |t²+2tu+u²| is not correct. It should be |t|²+2|tu|+|u|².

This makes (z²+2)² the wrong direction to take.

I've simply tried to fill in a couple of values for z.
With z = 1, we get |z+1|+|1-z+z²| = 3 > sqrt(7/2)

This is out-of-range, so the conclusion is that the original statement is false.
Qed.

 

[eumyang]

I've simply tried to fill in a couple of values for z.
With z = 1, we get |z+1|+|1-z+z²| = 3 > sqrt(7/2)

This is out-of-range, so the conclusion is that the original statement is false.
Qed.

This is incorrect. Looks like you missed post #8:

Mark44,thanks.I didn't notice it.Yes it's not sqrt(7/6) but actually 3sqrt(7/6).

So
\sqrt{\frac{7}{2}} \leq |z+1|+|1-z+z^2| \leq 3\sqrt{\frac{7}{6}}
is the correct inequality, and if you substitute in z = 1 the inequality holds.

 

[mikethehike]

get it in polar form z=e^(i*\alpha)

 

[I like Serena]

This is incorrect. Looks like you missed post #8:


So
\sqrt{\frac{7}{2}} \leq |z+1|+|1-z+z^2| \leq 3\sqrt{\frac{7}{6}}
is the correct inequality, and if you substitute in z = 1 the inequality holds.

True, I missed that one. Thanks for the correction.

I've gone on and filled in a few more numbers (since the equation is oddly difficult to solve).

arg z   (|z+1|+|1-z+z²|)²
π/3       3                        < 7/2
5π/6     10.56                     > 3² . 7/6 = 21/2

Again I find numbers that are out-of-range on both sides of the equation.
Did I miss something else?

Btw, I used Excel to find these numbers, which must be close to the minimum and maximum values.

[EDIT]Here's the graph I plotted

[/EDIT]

 

[lurflurf]

That should not be (5π/6,10.56), but rather (2Arccos(1/4),10.5625).

 

[I like Serena]

That should not be (5π/6,10.56), but rather (2Arccos(1/4),10.5625).

So how did you find it?

 

[I like Serena]

That should not be (5π/6,10.56), but rather (2Arccos(1/4),10.5625).

I've verified it with WolframMathematica (cool site )!

I'm concluding that the equation should be:

\sqrt 3 \leq |z+1|+|1-z+z^2| \leq \frac {13} {4}

I've solved this by taking the square, deriving it to phi, and setting the result to zero:
http://www.wolframalpha.com/input/?...sin(2x)+-+6+sin(3x)+/+sqrt(2+++2+cos(3x))+=+0

It finds the solutions 0 and 4 arctan sqrt(3/5).
Checking some more teaches me that 4 arctan sqrt(3/5) = 2 arccos 1/4 = 2 arctan sqrt 15.

Checking out pi and pi/3 learns that they are solutions as well.

So we have the following extrema:

\begin{matrix}<br /> z &amp; |z+1|+|1-z+z^2| \\<br /> \hline<br /> 1 &amp; 2 \\<br /> \pm \frac \pi 3 &amp; \sqrt 3 \\<br /> \pm 2 \arccos \frac 1 4 &amp; \frac {13} 4 \\<br /> \pi &amp; 3 \end{matrix}

 

[lurflurf]

let t=arg(z) with the usual convention -pi<t<=pi then by elementary trigonometry
|z+1|+|z^2-z+1|=a[2 cos(t/2)+(1/2)a]^2-(13/4)a
where
a=-1 when |t|=<pi/2
a= 1 when |t|>=pi/2

there will be local extrema when
2 cos(t/2)=0 ->t=0
a changes ->|t|=pi/3
2 cos(t/2)=1/2 ->|t|=2 Arccos(1/4)
2 cos(t/2)=1 ->t=pi

 

[I like Serena]

let t=arg(z) with the usual convention -pi<t<=pi then by elementary trigonometry
|z+1|+|z^2-z+1|=a[2 cos(t/2)+(1/2)a]^2-(13/4)a
where
a=-1 when |t|=<pi/2
a= 1 when |t|>=pi/2

there will be local extrema when
2 cos(t/2)=0 ->t=0
a changes ->|t|=pi/3
2 cos(t/2)=1/2 ->|t|=2 Arccos(1/4)
2 cos(t/2)=1 ->t=pi

Nice solution!
I feel stupid for not seeing it myself.