1. ## inequality

Hello everyone

I'd like to solve

$|\frac{4}{z-2}+1| \le 1$

I tried and I tried, but I failed. The good thing is, I found the solution, it should be $z \le 0$

Best regards,
Rapha

2. Originally Posted by Rapha
Hello everyone

I'd like to solve

$|\frac{4}{z-2}+1| \le 1$

I tried and I tried, but I failed. The good thing is, I found the solution, it should be $z \le 0$

Best regards,
Rapha
Hi

Should we suppose that z is a real and not a complex ?

If so $|\frac{4}{z-2}+1| \leq 1$

$-1 \leq \frac{4}{z-2}+1 \leq 1$

$-2 \leq \frac{4}{z-2} \leq 0$

$\frac{z-2}{4} \leq -\frac{1}{2}$

$z-2 \leq -2$

$z \leq 0$

3. Bon soir running-gag

Originally Posted by running-gag
Hi

Should we suppose that z is a real and not a complex ?

z is a complex. I didn 't know theres a difference.
How to solve it, if $z \in \mathbb{C}$ ?

Originally Posted by running-gag
If so $|\frac{4}{z-2}+1| \leq 1$

$-1 \leq \frac{4}{z-2}+1 \leq 1$

$-2 \leq \frac{4}{z-2} \leq 0$

$\frac{z-2}{4} \leq -\frac{1}{2}$

$z-2 \leq -2$

$z \leq 0$
Merci beaucoup.

4. Originally Posted by Rapha
Bon soir running-gag

z is a complex. I didn 't know theres a difference.
How to solve it, if $z \in \mathbb{C}$ ?

Merci beaucoup.
There is a difference
Order relation does not exist $\mathbb{C}$
If $z_1$ and $z_2 \in \mathbb{C}$ you cannot write $z_1 \leq z_2$

Therefore
Originally Posted by Rapha
The good thing is, I found the solution, it should be $z \le 0$
It is only possible if z is real, this is why I asked the question (also because "z" is the usual denomination for an unknown complex, otherwise you would have chosen "x")

5. Hello again!

Originally Posted by running-gag
There is a difference
Order relation does not exist $\mathbb{C}$
If $z_1$ and $z_2 \in \mathbb{C}$ you cannot write $z_1 \leq z_2$
I wasn't sure about it, because it is $|\frac{4}{z-2}+1| \leq 1$ and |z| = |x+iy| = x^2 + y^2

Alright, thank you very much!

Cordialement
Rapha

6. In this case :

$|\frac{4}{z-2}+1| \le 1$

$|\frac{z+2}{z-2}| \le 1$

$\frac{|z+2|}{|z-2|} \le 1$

$|z+2| \le |z-2|$

Let A be the point (-2), B (2) and M(z)

$|z+2|=AM$ and $|z-2|=BM$

$AM = BM$ means that M is on the "mediator line" of [AB] which is the y axis

$AM \le BM$ means that M is in the plane where $x \le 0$

7. [quote=running-gag;238875]In this case :

$|\frac{4}{z-2}+1| \le 1$

$|\frac{z+2}{z-2}| \le 1$

$\frac{|z+2|}{|z-2|} \le 1$

$|z+2| \le |z-2|$

Okay

Originally Posted by running-gag
Let A be the point (-2), B (2) and M(z)

$|z+2|=AM$ and $|z-2|=BM$

$AM = BM$ means that M is on the "mediator line" of [AB] which is the y axis

$AM \le BM$ means that M is in the plane where $y \le 0$
I'm sorry, I don't see your point. I even don't know how to graph that, but i imagine this is like an oval circle, isn't it?

8. No, look at the sketch below

$\overrightarrow{AM}$ affix is z-(-2)=z+2
Therefore AM = |z+2|

Please note that I have modified my previous post in x <=0 instead of y (I was mistaken)

9. Ah, okay, now I understand
Thank you, thank you, thank you

Best wishes
Rapha

10. Merry Christmas !
Bye