Is There a Connection Between Inequality and the Unit Disc?

Likemath2014
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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
 
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You need to clarify. For z = 1, the expression = 1, independent of λ.
 
In fact z is inside the disc that means |z|<1.
 
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)22
 
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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?
 
If z=0 and lamda=0.9999 then the LHS ~ 0.0001
 
Likemath2014 said:
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
 
The statement the OP is trying to prove is not true. See #6. z=0 and lamda = 0.9999 is a counterexample.
 
FactChecker said:
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.
 
  • #10
mathman said:
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)
 
  • #11
FactChecker said:
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.
 
  • #12
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.
 
  • #13
Likemath2014 said:
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
 
  • #15
mathman said:
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 :confused:.

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

a^2+b^2+2a(\lambda-1)+(\lambda-1)^2&lt;\lambda^2.
How could that mean 2a(λ-1) < 2(λ-1)?
Thx.
 
  • #16
Since a^2+b^2 &lt; 1, |a| &lt; 1
 
  • #17
How could that help :redface:.
 
  • #18
Likemath2014 said:
How could that help :redface:.

Grade school arithmetic! If |a| \ge 1, then a^2 \ge |a| \ge 1 contradicting a^2 + b^2 &lt; 1
 
  • #19
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)+(<br /> \lambda-1)^2\leq\lambda^2

implies

2a(λ-1) < 2(λ-1)?
 
  • #20
Likemath2014 said:
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)+(<br /> \lambda-1)^2\leq\lambda^2

implies

2a(λ-1) < 2(λ-1)?
By itself it doesn't. |z| < 1 is the needed condition.
 
  • #21
Yes, the question is how is that?
 
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