# Inequality in the unit disc

1. Aug 22, 2014

### 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

Last edited: Aug 22, 2014
2. Aug 22, 2014

### mathman

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

3. Aug 27, 2014

### Likemath2014

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

4. Aug 27, 2014

### 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)22

5. Aug 27, 2014

### 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?

6. Aug 30, 2014

### FactChecker

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

7. Aug 31, 2014

### mathman

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

8. Aug 31, 2014

### FactChecker

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

9. Sep 1, 2014

### mathman

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

10. Sep 1, 2014

### FactChecker

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

11. Sep 2, 2014

### mathman

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

12. Sep 2, 2014

### 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.

13. Sep 2, 2014

### FactChecker

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

14. Sep 2, 2014

### Likemath2014

15. Sep 3, 2014

### Likemath2014

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.

16. Sep 4, 2014

### mathman

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

17. Sep 4, 2014

### Likemath2014

How could that help .

18. Sep 5, 2014

### mathman

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

19. Sep 9, 2014

### 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)?

20. Sep 9, 2014

### mathman

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

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