Is There a Connection Between Inequality and the Unit Disc?

[Likemath2014]

hi there,

I am trying to prove the following inequality:
let $$z\in \mathbb{D}$$ then

$$\left| \frac{z}{\lambda} +1-\frac{1}{\lambda}\right|<1$$ if and only if $$\lambda\geq1.$$

The direction if $$\lambda>1$$ is pretty easy, but I am wondering about the other direction.

Thanks in advance

 

[mathman]

You need to clarify. For z = 1, the expression = 1, independent of λ.

 

[Likemath2014]

In fact $$z$$ is inside the disc that means $$|z|<1$$.

 

[mathman]

z = a+bi. Restate question. |a+bi+λ-1|² < λ²
L.H.S. = (a+λ-1)²+b²=|z|²+2a(λ-1)+(λ-1)²<1+2(λ-1)+(λ-1)²=λ²

 

[Likemath2014]

thank, but how could that imply that lambda is bigger than one? On the other hand how did you get the last inequality, why you consider a is positive?

 

[FactChecker]

If z=0 and lamda=0.9999 then the LHS ~ 0.0001

 

[mathman]

thank, but how could that imply that lambda is bigger than one? On the other hand how did you get the last inequality, why you consider a is positive?

2a(λ-1) < 2(λ-1) requires λ > 1 for a > 0. Also |a| < 1

 

[FactChecker]

The statement the OP is trying to prove is not true. See #6. z=0 and lamda = 0.9999 is a counterexample.

 

[mathman]

The statement the OP is trying to prove is not true. See #6. z=0 and lamda = 0.9999 is a counterexample.

Your counterexample is wrong. λ ≥ 1 is a condition.

 

[FactChecker]

Your counterexample is wrong. λ ≥ 1 is a condition.

The OP said "If and only if". lambda is less than 1 but the equation is much less than 1 (nearly 0)

 

[mathman]

The OP said "If and only if". lambda is less than 1 but the equation is much less than 1 (nearly 0)

You misread the op. λ≥1 is the condition. For λ < 1 the expression does not hold for all z, with |z| < 1.

 

[Likemath2014]

Maybe it was not clear in the question that the inequality is for all z in the disc, that was my fault, I am sorry. Thank you very much mathman for the solution.

 

[FactChecker]

Maybe it was not clear in the question that the inequality is for all z in the disc, that was my fault, I am sorry. Thank you very much mathman for the solution.

Oh. Well, that's different. Never mind. https://www.youtube.com/watch?v=V3FnpaWQJO0

 

[Likemath2014]

Oh. Well, that's different. Never mind. https://www.youtube.com/watch?v=V3FnpaWQJO0

 

[Likemath2014]

2a(λ-1) < 2(λ-1) requires λ > 1 for a > 0. Also |a| < 1

I thought I got it, but it seems not yet .

We will start like that,
let
$$|z+\lambda-1|^2 <|\lambda|^2$$, fro all z in the disc. Let $$z=a+ib$$, hence

$$a^2+b^2+2a(\lambda-1)+(\lambda-1)^2<\lambda^2.$$
How could that mean 2a(λ-1) < 2(λ-1)?
Thx.

 

[mathman]

Since $a^2+b^2 < 1, |a| < 1$

 

[Likemath2014]

How could that help .

 

[mathman]

How could that help .

Grade school arithmetic! If $|a| \ge 1$, then $a^2 \ge |a| \ge 1$ contradicting $a^2 + b^2 < 1$

 

[Likemath2014]

of course that is a grade school arithmetic, but it was not my question.

My question is:

how
$$a^2+b^2 +2a( \lambda -1)+( \lambda-1)^2\leq\lambda^2$$

implies

2a(λ-1) < 2(λ-1)?

 

[mathman]

of course that is a grade school arithmetic, but it was not my question.

My question is:

how
$$a^2+b^2 +2a( \lambda -1)+( \lambda-1)^2\leq\lambda^2$$

implies

2a(λ-1) < 2(λ-1)?

By itself it doesn't. |z| < 1 is the needed condition.

 

[Likemath2014]

Yes, the question is how is that?