inequality

**Rapha** · December 19th 2008, 07:03 AM

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 $z \le 0$

Can someone please help me?

Thank you in advance!

Best regards,
Rapha

**running-gag** · December 19th 2008, 07:35 AM

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 $z \le 0$

Can someone please help me?

Thank you in advance!

Best regards,
Rapha

Hi

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

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

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

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

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

$z-2 \leq -2$

$z \leq 0$

**Rapha** · December 19th 2008, 07:44 AM

Bon soir running-gag

Originally Posted by running-gag

Hi

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

Good thing you ask that.

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.

**running-gag** · December 19th 2008, 08:00 AM

Originally Posted by Rapha

Bon soir running-gag

Good thing you ask that.

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")

**Rapha** · December 19th 2008, 08:04 AM

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

**running-gag** · December 19th 2008, 08:20 AM

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 $x \le 0$

**Rapha** · December 19th 2008, 08:30 AM

[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?

**running-gag** · December 19th 2008, 08:40 AM

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)

**Rapha** · December 19th 2008, 08:50 AM

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

Best wishes
Rapha

**running-gag** · December 19th 2008, 08:55 AM

Merry Christmas !
Bye

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