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Thread: inequality

  1. #1
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    inequality

    Hello everyone

    I'd like to solve

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

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

    Can someone please help me?

    Thank you in advance!

    Best regards,
    Rapha
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  2. #2
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    Quote Originally Posted by Rapha View Post
    Hello everyone

    I'd like to solve

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

    I tried and I tried, but I failed. The good thing is, I found the solution, it should be $\displaystyle 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 $\displaystyle |\frac{4}{z-2}+1| \leq 1$

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

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

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

    $\displaystyle z-2 \leq -2$

    $\displaystyle z \leq 0$
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  3. #3
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    Bon soir running-gag

    Quote Originally Posted by running-gag View Post
    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 $\displaystyle z \in \mathbb{C}$ ?

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

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

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

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

    $\displaystyle z-2 \leq -2$

    $\displaystyle z \leq 0$
    Merci beaucoup.
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  4. #4
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    Quote Originally Posted by Rapha View Post
    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 $\displaystyle z \in \mathbb{C}$ ?



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

    Therefore
    Quote Originally Posted by Rapha View Post
    The good thing is, I found the solution, it should be $\displaystyle 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")
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  5. #5
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    Hello again!

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

    Alright, thank you very much!

    Cordialement
    Rapha
    Last edited by Rapha; Dec 19th 2008 at 08:19 AM.
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  6. #6
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    In this case :

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

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

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

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

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

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

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

    $\displaystyle AM \le BM$ means that M is in the plane where $\displaystyle x \le 0$
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  7. #7
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    [quote=running-gag;238875]In this case :

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

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

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

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

    Okay


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

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

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

    $\displaystyle AM \le BM$ means that M is in the plane where $\displaystyle y \le 0$
    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?
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  8. #8
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    No, look at the sketch below

    $\displaystyle \overrightarrow{AM}$ 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)

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

    Best wishes
    Rapha
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  10. #10
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    Merry Christmas !
    Bye
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